GROUP (A)–CLASS WORK PROBLEMS
GROUP (A)–HOME WORK PROBLEMS
Q-1) Find the degree and order of following
differential equation
Q-1) Find the degree and order of following
differential equation.
2
3/2
3
 d 2y 
 dy 
x
+
i) 
 =e
2 
dx
dx




 d 3y 
i) 
3 
 dx 
Ans. Order = 2, Degree = 2
Ans. Squaring both sides
=x
3
dy
dx
dy
 dy 
+r 1+ 
ii) y = x

dx
 dx 
 d 3y 
 3
 dx 
Ans. Squaring both sides
Order : 3, Degree : 3
2
dy 

 y – x dx 


2
  dy 2 
= r 1 + 
 
  dx  
2
Order = 1, Degree = 2
dy

 y – x dx
iii) 
dy

dx


  dy 2
=

  dx 

dy

 y – x dx

 dy
 dx
 dy 
= 

 dx 
(
1+
dx
)
2
1
(
2
dy
dx
)
 d 2y 
= 2 
 dx 
3
2
dy 
dy

+x
= x2
iii)  y +

dx 
dx

2
Ans. y 2 + 2y
4
dy
dy  dy 
+x
+
= x2
dx
dx  dx 
Order : 1, Degree : 2
3
dy
iv) e dx +
Order = 1, Degree = 3
d 2y 3
 dy 
= 1– 
iv)

2
dx
 dx 
dy
2
3



Order : 2, Degree : 3

2

 dy 
 = 

 dx 


dy
dx
1+
 d 2y
=
2
 dx
1
2
Ans. Squaring
Ans. The given D. E. is
⇒ y–x
ii)
 dy 
x2 

 dx 
=
dy
=x
dx
Ans. By definition of degree and order
Degree : Not defined; Order : 1
Ans. On cubing both sides
3
 d 2y 
 2
 dx 
∴
 dy 
= 1+ 

 dx 
x3
v)
2x 2
Order = 2, Degree = 3
5x
 d 2y
v) 
2
 dx
y2
dy
3y
dx
 d 2 y  dy
2 y
+
2
 dx
 dx
3
0 =0



2

 0

2

 dy 
 + cos 
=0
 dx 

By definition of degree and order Degree :
not defined; Order : 2
2


d 2y
2  dy 
Ans. x3(0) – y3(0) + 3 4x 2y
+
4
x
 dx  
2
dx

 

– 15xy
dy
=0
dx
Differential Equation
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2
2
∴
d 2y
dy
 dy 
12x y
+12x 2 
=0
 – 15xy
2
dx
dx
 dx 
2
2
i.e 4xy
∴
d 2y
dy
 dy 
+ 4x 
 – 5y dx = 0
dx
dx 2


Q-3)
xy = Aex + Be–x + x2
Ans. The given equation is
xy = Aex + Be–x + x2
xy = x2 + Aex + Be–x
As there are two arbitary constants we have
to differentiate twice
Differentiating both sides twice w. r. t. x
Order = 2, Degree = 1
GROUP (B)- CLASS WORK PROBLEMS
we get
x
dy
+ y – 2x = Aex – Be–x
dx
Find the differential equation by eliminating the
arbitary constants from the following equations.
Q-1) y = Acos 4x + Bsin 4x
 d 2y dy  dy
+
– 2 = Aex + Be–x
x
+
2
dx
dx
dx


Ans. The given equation
is y = A cos 4x + B sin 4x
x
d 2y
dy
+2
– 2 = xy – x2
2
dx
dx
x
d 2y
dy
+2
– 2 = x(y – x)
2
dx
dx
...(i)
As there are two arbitary constant, we have
to differentiate twice.
Differentiating w. r. t. x we get,
This is the required D.E.
dy
= – 4A sin 4x + 4B cos 4x
dx
∴
d 2y
= – 16A cos 4x – 16B sin 4x
dx 2
∴
d 2y
= – 16(A cos 4x + B sin 4x)
dx 2
∴
d 2y
= – 16y
dx 2
∴
d 2y
+ 16y = 0 is required D.E.
dx 2
...[By (i)]
Q-2) y = c1emx + c2e–mx
Ans. The given equation is
y = c1emx + c2e–mx
... from (i)
Q-4) xy = ae3x + be –2x
Ans. The given equation is
xy = ae3x + be – 2x
As there are two arbitary constant we have
to differentiate twice
Differentiating w. r. t. x. we get
x
dy
+y
dx
= 3ae3x – 2be
– 2x
...(ii)
Again Diff w.r.t x
x
d 2y dy dy
+
+
= 9ae3x + 4be
dx 2 dx dx
– 2x
d 2y
dy
+2
= 9ae3x + 4be
2
dx
dx
– 2x
…(i)
As there are two arbitary constant, we have
to differentiate twice
…(i)
i.e. x
...(iii)
Differentiating twice w. r. t. x., we get
dy
dx
= mc1emx – mc2e–mx
∴
dy
dx
= m(c1emx – c2e–mx)
∴
dy
dx
= m2(c1emx + c2e–mx)
2
= my
∴
d 2y
= m 2y ;
dx 2
d 2y
– m 2y = 0
dx 2
is the required DE
Differential Equation
xy
dy
x
+y
dx
d 2y
dy
x
+2
2
dx
dx
1
1
3 –2 = 0
9
4
 dy
d 2y
dy 
xy [12 +18] – 1 4x
+ 4y + 2x
+4

2
dx 
dx
 dx
 dy
d 2y
dy 
+1 9x
+ 9y – 3x
–6
=0
2
dx 
dx
 dx
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3
dy
d 2y
dy
30xy – 4x
– 4y – 2x
–4
dx
dx
dx 2
2
dy
d y
dy
+ 9x
+ 9y – 3x
–6
=0
dx
dx
dx 2
Q-6) y = (c1 + c2x)e2x
Ans. The given equation is y = (c1 + c2x)e2x
As there are two arbitary constants we have
∴
d 2y
dy
dy
–5x
+ 5x
– 10
+ 30xy + 5y = 0
2
dx
dx
dx
2
x
d y
dy
dy
–x
+2
– 6xy – y = 0
2
dx
dx
dx
2
x
d y
dy
– ( x – 2)
– ( 6x +1) y = 0
dx
dx 2
Differentiating again w. r. t. x. we get,
e –2 x
∴
 dy

e –2 x 
– 2y 
dx


d 2y
dx 2
 d 2y
dy 

–2 x  dy
– 2y  = 0
e –2x  2 – 2
 – 2e

dx 
 dx

 dx
∴
 d 2y

dy
dy
e –2 x  2 – 2
–2
+ 4y  = 0
dx
dx
 dx

∴
d 2y
dy
–4
+ 4y = 0
dx
dx 2
...(i)
dy
=0
dx
This is the required D.E.
...(ii)
 d 2y dy dy 
Diff. A + B y
+
.
=0
2
dx dx 
 dx
 d 2y  dy 2 
A + B y
+
 =0
2
 dx  
 dx
Q-7)
y2
dy
y
dx
x
∴
1
y
∴
 dy 
ye–3x (–3) + e–3x 

 dx 
∴
 dy

e –3x 
– 3y  = –2Ae–2x + 4Be–4x
 dx

∴
2
0
 dy

ex 
– 3y 
 dx

= –2Ae2x + 4B
Differentiating w. r. t. x. we get,
x2(0 – 0) – y2 (0 – 0) + 1
∴
  d 2y dy 2
 dy  
x  y
+
=0
 –y
2
dx 
dx  
  dx



 d 2y
dy 
x
ex  2 – 3
 +e
dx
dx


∴
 d 2y

dy
ex  2 – 2
– 3y 
dx
 dx

2
xy
= – 2Ae–2x – 4Be–4x
Divide by e–4x
0 =0
d y  dy 
+

dx 2  dx 
have to differentiate thrice
e–3xy = Ae–2x + Be–4x + C
Differentiating again w. r. t. x., we get,
...(iii)
1
2
y = Aex + Be–x + Ce3x
Ans. The given equation is y = Aex + Be–x + Ce3x
As there are three arbitary constants, we
Equation (i),(ii) and (iii) are consistent
x2
= c2
∴
dy
A ( 2x ) + B ( 2y )
=0
dx
Ax + By
= c2
Differentiating w. r. t. x. we get,
Q-5) Ax2 + By2 = 1
Ans. As there are two arbitary constant we have to
2
2
Diff. Ax + By =1
dy
– 2e –2 x y
dx
∴
This is the required D.E
find
to differentiate twice
e –2x y = c1 + c2x
d 2y
 dy 
 dy 
+x
 – y  dx  = 0
dx 2
 dx 


 dy

 dx – 3y 


= –4Ae2x
= –4Ae2x
Divide by e2x
∴
 d 2y

dy
e –x  2 – 2
– 3y 
dx
 dx

= –4A
Differentiating w. r. t. x., we get,
Differential Equation
Mahesh Tutorials Science
4
∴
e
–x
Put (ii) in (i)
 d 3y
d 2y
dy 
 3 –2 2 –3
+
dx 
dx
 dx
y =
 p 
A

 2A 
y =
 d 3y

d 2y dy
+ 3y  = 0
e –x  3 – 3 2 –
dx
dx
dx


p2 A
4A 2
y =
p2
4A
d 3y
d 2y dy
–
3
–
+ 3y = 0
dx 3
dx 2 dx
A =
p2
4y
 d 2y

dy
– 3y  –e – x = 0
 2 –2
dx
 dx

(
∴
)
2
is the required D.E.
x–
Q-8) (x – a)2 + y2 = a2
Ans. The given equation is
(x – a)2 + y2 = a2
...(i)
=
4xy – p 2
4y
As there is one arbitary constant we have to
differentiate once
p
 p2 
2

 4y 
2py
p2
=
p (4xy – p2) = 8y2
Differentiate w.r.t.x we get
dy
2(x – a) + 2y
dx
p2
4y
4pxy – p 3 = 8y2
p 3 – 4xyp + 8y2 = 0
= 0
3
dy
(x – a) = –y
dx
∴
 dy 
 dy 
2
 dx  – 4xy  dx  + 8y = 0




...(ii)
is the required D.E
dy
a=x+y
dx
...(iii)
substituting these values in (i) we get
Q-10)Form the differential equation of all
parabolas whose axis is x-axis.
Ans. Equation of such parabola is
2
2
dy 

dy 

2
–
y
+
y
= x +y

dx 
dx 


2
∴
dy
 dy 
 dy 
2
2
y2 
+ y2 
 + y = x + 2xy

dx
 dx 
 dx 
∴
y2 = x2 + 2xy
2xy
y2 = 4a(x – h) where vertex is (h, 0)
y2 = 4a (x – h)
...(i)
2
Differentiating w.r.t.x, again,
dy
dx
2y
dy
= 4a
dx
 d 2y  dy 2 
2 y
+
 =0
2
 dx  
 dx
dy
+ x2 – y2 = 0
dx
is the required DE
2
∴
Q-9) y = A (x – A)2
Ans. y = A (x – A)2
= A[2 (x – A)]
x–A =
1 dy
2A dx
x–A =
1
dy 

p  where p =

2A 
dx 
Differential Equation
d 2y  dy 
+
 = 0 is the required D.E.
dx 2  dx 
…(i)
As there is one arbitary constant, we have
to differentiate once
dy
dx
y
Q-11) Form the differential equation of all circles
having their centres on the line y = 10 and
touching x-axis.
Ans. Let the centre of circles be (h, 10) and as the
circle touches x-axis, so the radius of circles
is 10 units.
Equation of family of circles is
…(ii)
(x – h)2 + (y – 10)2 = (10)2
...(i)
Differentiating w.r.t.x, we get,
Mahesh Tutorials Science
2(x – h) + 2(y – 10)
(x – h) + (y – 10)
5
dy
=0
dx
dy
=0
dx
...(ii)
From (ii) x – h = –(y – 10)
 dy

d
d
3x
– 3b
e –3 x
 x dx + y  = 3a dx e
dx


d
dx
x
dy
dx
(
)
(
)
d  dy  dy d
dy
+
.
(x ) +
dx  dx  dx dx
dx
= 3ae3x
d
d
(3x) – 3be–3x
(–3x)
dx
dx
Squaring on both sides, we get
 dy 
(x – h)2 = –(y – 10)2 

 dx 
x
2
= 9(ae 3x + be –3x)
2
Putting this value of (x – h) in equation (i)
we get
2
 dy 
2
2
(y – 10)2 
 + (y – 10) = (10)
 dx 
... (i)
... (ii)
dy
e + c.e
=0
dx
... (iii)
Put (ii) in (iii)
... (i)
–3x
Q-2) xy = ae + be
Ans. As there is two arbitary constant we have to find
d 2y
⇒ xy = ae 3 x + be –3 x
dx 2
d
d
d
e 3x + b
e –3 x
( xy ) = a
dx
dx
dx
(
)
x
dy
dx
+y
dx
dx
x
dy
+ y = 3ae3x – 3be–3x
dx
ae2x –1(2) + be –2x – 1(–2)
dy
= 2ae2x –1 – 2be –2x
dx
–1
d  dy 
dx  dx 
d
d
(2x – 1) – 2be –2x – 1
(–2x –1)
dx
dx
d 2y
= 2ae2x –1(2) – 2be –2x – 1(–2)
dx 2
dy
=0
dx
diff both sides
dy
+0 =
dx
= 2ae2x –1
 y dy
=0
 .e
dx

ex + (1 – ex )
3x
–1
Differentiating again w. r. t. x
y
1 – ex
ex + 
y
 e
–1
Diff w. r. t. x
Differentiating w.r.t.x.
x
–1
y + 1 = ae2x –1 + be –2x
Q-1) ex + cey = 1
Ans. ex + cey = 1
c
)
d 2y
dx 2
GROUP (B)- HOME WORK PROBLEMS
∴
(
Ans. As there is two artibary constants we have
to find
is the required D.E
1 – ex
=
ey
d 2y 2dy
+
– 9 ae 3 x + be –3 x = 0
dx 2
dx
Q-3) y = ae2x – 1 + be –2x
 dy 

+1 = 100
(y – 10)2 


 dx 
cey = 1 – ex
= x
This is the required D.E
2
∴
d 2y dy dy
+
+
= 3ae3x (3) – 3be–3x (–3)
dx 2 dx dx
(
d 2y
= 4[ae2x –1 + be –2x
dx 2
–1
]
d 2y
= 4(y + 1)
dx 2
d 2y
– 4y – 4 = 0
dx 2
)
Q-4) y = ye2x [A cos x + B sin x]
Ans. y = e2x [A cos x + B sin x]
3x
–3 x
= a 3e  + b  –3e 
... (i)
As there are two arbitary constants we have
... (ii)
to find
d 2y
dx 2
d
ye 2 x
dx
(
)
= A
d
d
(cos x) + B
(sin x)
dx
dx
Differential Equation
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6
y
d 2x
dy
e + e 2x
= – A sin x + B cos x
dx
dx
y.e2x(2) + e2x
dy
=
dx
– A sin x + B cos x
6y – 9
dy
= – A sin x + B cos x
dx
2ye2x + e2x
Diff. w.r.t. x
2ye2x(2) + e2x (2)
dy
d 2y dy 2x
+
+ e2x
e (2)
dx
dx 2 dx
dy
d 2y
dy
d 2y
+3 2 +4
–2 2 =0
dx
dx
dx
dx
dy
d 2y
=5
+ 6y = 0
dx
dx 2
This is required D.E
Q-6) y = a cos (logx) + b sin(log x)
Ans. As there are two arbitary constants we have
= – A sin x – B sin x
2
4ye2x + 2e2x
 dy
 dy
d 2y 
d 2y 
y [18 – 12] – 1 9
– 3 2  +1 4
– 2 2 =0
dx 
dx 
 dx
 dx
dy
d y
dy 2x
+ e2x
e
+2
2
dx
dx
dx
= – A cos x – B sin x
d 2y
to find
dx 2
y = a cos (log x) + b sin (log x)
2
4ye2x + 2e2x
dy
d y
dy 2x
+ e2x
+2
e
2
dx
dx
dx
dy
1
1
= – a sin (log x) ×
+ b cos (log x) ×
dx
x
x
= – (A cos x + B sin x)
4ye
e
2x
2x
+ 4e
2x
dy
d 2y dy
+ e2x
= –ye2x
+
dx
dx 2 dx
d 2y
dy
+ 4e 2 x
+ 4ye2x + ye2x = 0
2
dx
dx
d 2y
dy
+4
+ 4y + y = 0
2
dx
dx
Q-5) y = Ae2x + Be3x
Ans. As there are two arbitary constant we have
d 2y
dx 2
y = Ae2x + Be3x
... (i)
d 2y
= 4Ae2x + 9Be3x
dx 2
... (ii)
Differentiating w.r.t.x
x
d 2y dy
+
(1)
dx 2 dx
1
1
– b sin ( log x ) ×
x
x
x
d 2y dy –1
+
=
[a cos(log x) + b sin(log x)]
x
dx 2 dx
x
d 2y dy –y
+
=
x
dx 2 dx
x2
diff. w.r.t.x
dy
= 2Ae2x + 3Be3x
dx
dy
= – a sin (log x) + b cos (log x)
dx
= – a cos (log x)
d 2y
dy
+4
+ 5y = 0
2
dx
dx
to find
x
... (i)
d 2y
dy
+x
+y = 0
dx
dx 2
This is required D.E
2
Q-7) y = Ax + Bx
... (ii)
... (iii)
Equation (i),(ii) & (iii) are consistent
Ans. As there is two arbitary constant we have
to find
d 2y
dx 2
y = A x2 + B x2
Differentiating w.r.t.x
y
dy
dx
d 2y
dx 2
1 1
2 3 =0
4 9
dy
= 2Ax + B (1)
dx
Differentiating w.r.t.x
d 2y
= 2Ax + B (0)
dx 2
Equation (i), (ii) & (iii) are consistent
Differential Equation
... (ii)
... (iii)
Mahesh Tutorials Science
Again, diffferentiating w.r.t. x, we get
x2
y
dy
dx
d 2y
dx 2
7
x
2x
0+
1 =0
2
0
∴

 dy
d 2y 
d 2y 
y ( 0 – 2) − x 2  0 –
+ x 2
–2 2 =0
2 
dx 
dx 

 dx
d 2y
=0
dx 2
d 2y
=0
dx 2
This is required D.E
Q-10)The equation of the ellipse is given as
(2x
x2
2
– x2
)
d 2y
dy
– 2x
+ 2y = 0
2
dx
dx
x2 y2
+
= k . F orm
th e
differen tial
36 16
equation of the ellipse whose length of
d 2y
dy
– 2x
+ 2y = 0
dx
dx 2
major and minor axes are half the lengths
of the given ellipse respectively.
This is required D.E
Ans. Conside r e quation of the given ellipse
2
x2 y2
+
=k
36 16
Q-8) (y – a ) = 4(x – b )
Ans. The given equation is (y – a )2 = 4(x – b )
... (i)
Differentiating twice w.r.t. x, we get,
dy
dy
2(y – a )
= 4 , i.e., (y – a )
=2
dx
dx
∴
x2
(
6 k
2
+
) (
y2
4 k
)
2
=1
... (ii)
This is of the form
2
and (y – a )
d 2y  dy 
+
 =0
dx 2  dx 
From (ii), y – a =
... (iii)
∴
a = 6 k and b = 4 k
Let the equation of required ellipse be
2
dy / dx
x 2 y2
+
=1
A2 B 2
Substituting this value in (iii), we get,
According to given condition,
2
2
d 2y  dy 
.
+
=0
dy / dx dx 2  dx 
A=
3
∴2
d 2y  dy 
+
 =0
dx 2  dx 
This is the required D.E.
b
a
and B =
2
2
∴
A = 3 k and B = 2 k
∴
The equation of required ellipse is
x 2 y2
+
=1
9k 4k
Q-9) Find the differential equation of the family
of lines where length of the normal from
origin is p and the inclination of the normal
is α.
∴
x 2 y2
+
=k
9
4
Differentiating w.r.t. x, we get
Ans. Equation of line is x cos α + y sin α = p
Dividing throughout by ‘sin α’ , we get
x cot α + y =
2x
+
9
p
sin α
Differentiating w.r.t. x, we get
cot α(1) +
dy
=0
dx
x 2 y2
+
=1
a 2 b2
∴
x
+
9
dy
dx = 0
4
2y
y
dy
dx = 0
4
Differential Equation
Mahesh Tutorials Science
8
dy
=0
dx
∴
4x + 9y
∴
dy –4x
=
dx
9y
∴
∴
dy
= sec2 x
dx
y sec x tan x + sec x.
Dividing by sec x, we get
dy
= sec x ( ∵ sec x ≠ 0)
dx
y tan x +
 dy 
9y 
 = –4x
 dx 
∴
y sec x = tan x + c
This is the required general solution of the
 dy 
4x + 9y 
=0
 dx 
given D.E.
Q-3) Verify that x2 + y2 = r2 is a solution of the
differential equation.
This is required D.E
GROUP (C)–CLASS WORK PROBLEMS
2
dy
 dy 
y=x
+r 1+ 
 .
dx
 dx 
Q-1) Show that y = aex + be–2x is the general
Ans. x2 + y2 = r2
solution of the differential equation
...(i)
Diff. w.r.t.x, we get
2
d y dy
+
– 2y = 0 .
dx 2 dx
2x + 2y.
Ans. Consider y = aex + be–2x
...(i)
Differentiating w.r.t.x. we get
dy
dx
= aex – 2be–2x
∴
...(ii)
L.H.S =
dy
 dy 
+ r 1+ 

dx
 dx 
...(iii)
 x
 x
= x .  –  + 1+  – 
 y
 y
d y dy
+
– 2y
dx 2 dx
(
...(ii)
2
RHS = x
2
∴
–x
dy
=
y
dx
Substituting in the RHS of the given D.E. we get
Differentiating w.r.t.x. we get
d 2y
= aex + 4be–2x
dx 2
dy
=0
dx
) (
)
)
x
–2 x
+ ae x – 2be –2 x
= ae + 4be
(
– 2 ae x + be − x
= 0 = R.H.S.
2
–x 2
r2
+r
y
y2
=
x2 r 2
+
y
y
=
–
=
r 2 – x2
y
=
y2
y
....By (i)
y = aex + be–2x is solution of given D.E
It contains two arbitary constants. Also order
of given differential equation is 2. Hence,
y = aex + be–2x is the general solution of the
given D.E.
= y = L.H.S.
Q-2) Show that y sec x = tan x + c is general
solution of the differential equation
dy
+ y tan x = sec x .
dx
Ans. y sec x = tan x + c
Diff.w.r.t. x, we get
Differential Equation
....By (ii)
This shows that x2 + y2 = r2
is a solution of the D.E.
dy
 dy 
+ r 1+ 
y= x

dx
 dx 
2
Mahesh Tutorials Science
9
Q-4) (1 + x2) dy – xy dx = 0
Ans. (1 + x2) dy – xy dx = 0
Q-6)
dy
= 1 + x + y + xy
dx
Ans.
dy
dx
Dividing by (1 + x2)y, we get
dy
x
–
dx = 0
y
1+ x2
= (1 + x)(1 + y)
Integrating we get,
∫
1
dy –
y
∫
1
1
log y –
dy
dx
x
dx = c1
1+ x2
2x
∫ y dy – 2 ∫ 1 + x
2
(
log y – log 1 + x 2

y
log 

2
 1 + x
(
)
1
)
2
Integrating,
where
= log c
12
(1+ x )
2
dy
∫ (1 + x ) dx + c
∴
∫ 1+ y
∴
log 1+ y = x +
=
x2
+c
2
This is the required general solution

 = log c
12 

)
Q-7)
y
= (1 + x)(1 + y)
1
dy = (1 + x)dx
1+ y
dx = c1
1
log 1 + x 2 = log c
2
(
= 1(1 + x) + y (1 + x)
 e 3 x – e –3 x 
dy
= y  3x
–3 x 
dx
e +e

=c
 e 3 x – e –3 x
1
dy =  3 x
–3 x
y
e +e
Ans. ∴
y = c 1 + x 2 is the required general solution.

 dx

Integrating
Q-5)
dy 1 + y 2
=
dx 1 + x 2
∴
∫
1
dy
y
∫
=
e 3 x – e –3 x
dx
e 3 x + e –3 x
...(i)
Put e3x + e–3x = t Then
2
Ans. We have
dy
1+ y 2
dy 1+ y
=
dx 1+ x 2
dx
=
1+ x2
3(e3x – e –3x)dx =
dt
∴
(e3x – e –3x)dx
dt
3
∴
from (i)
dx
dy
–
=0
1+ x 2 1+ y 2
1
=
1 dt
∫ y dy = ∫ t . 3
on integrating, we have
log|y|
∴
3log|y| =
tan–1 x – tan–1 y = c
∴
log|y3| = log |c13 (e3x + e–3x)|
This is the required general solution.
∴
y3 = c(e3x + e–3x), when c = c13
∫
∴
∫
=
1
log|t|+ log c1
3
∴
dx
dy
–
=0
1+ x 2
1 + y2
log |e3x + e–3x| + 3logc1
This is the required general solution
Differential Equation
Mahesh Tutorials Science
10
 dy 
Q-8) log 
 = 2x + 3y
 dx 
=
y sin 2y
–
2
 dy 
Ans. log 
 = 2x + 3y
 dx 
=
y sin 2y cos 2y
+
2
4
∫
sin 2y
dy
2
Similarly,
dy
dx
=e
∴
dy
dx
= e2x.e 3y
∴
e–3ydy
= e2x.dx
∴
∫e
∴
2x + 3y
I2 =
2x
dx –
∫e
–3 y
∫ x cos 2x dx =
x sin 2x cos 2x
+
2
4
From (i)
∴
dy = c1
e 2 x e –3 y
+
= c1
2
3
x 2 + y 2 y sin 2y cos 2y
+
+
+
2
2
4
x sin 2x cos 2x
+
= c1
2
4
(
)
2 x 2 + y 2 + 2 ( x sin 2x + y sin 2y )
+ cos 2x + cos 2y = c
∴ 3e2x + 2e –3y = 6c1
∴ 3e2x + 2e –3y = c
where c = 4c1
where c = 6c1
This is the required general solution
This is the required general solution
Q-10)
dy
π
sin x = y log y, when x = , y = e
dx
2
Ans.
dy
sin x = y log y,
dx
∴
dy
y log y
∴
1
y
dy
log y
Q-9) x–1 cos2 y dy + y–1 cos2 x dx = 0
Ans. x–1 cos2 y dy + y–1 cos2 x dx = 0
∴
cos 2 y
cos 2 x
dy +
dx = 0
x
y
∴
y cos2 y dy + x cos2 x dx = 0
∴
 1 + cos 2y 
 1 + cos 2x
y
 dy + x 
2
2



∴
y dy + y cos 2y dy + xdx + x cos 2x dx = 0

 dx = 0

∫
∫
∫ xdx +∫ x cos 2x dx = c
∴
y2
x2
+ y cos 2y dy +
+
2
2
∫ x cos 2x dx = c
∴
x 2 + y2
+ y cos 2y dy +
2
∫ x cos 2x dx = c
∫
∫
1
1
∫
∴
d
( log y ) =
dy
⇒
x 
log(log y) – log tan   = log c
2
∴

log y
log 

x
 tan 2
∴
x 
log y = C tan   is the required general
2
...(i)
∫
I1 =
∫ y cos 2y dy
Differential Equation
1
y
dy –
log y
∴
I 1 = y cos 2y dy
Integrating by parts we get
= cosec x dx
Integrating
Integrating we have
y dy + y cos 2y dy +
dx
sin x
=
∫ cosec x dx = log c
( )
solution
1
y

 = log C


Mahesh Tutorials Science
when x =
11
π
; y = e, we have
2
π
log e = C tan  
4
∴
dy
=
x 25 – x 2 dx
∴
∫ dy
=
1
2
25 – x 2 ( 2x ) dx + c
Put 25 – x2 = t ⇒ ∴ 2xdx = – dt
1=C
∴
∫
1
2
∫
t ( –dt ) + C = –
The particular solution is
∴
y=
π
log y = tan  
2
∴
y= –
∴
3y + (25 – x2)3/2 = 3C
Q-11)
dy
– x2 = x2 y if x = 0, when y = 2
dx
∴
∴
dy
(1 + y )
1
. (25 – x2)3/2 + C
3
when x = 0, y = 1/3 we have
1
3   + (25 – 0)3/2 = 3C
3
dy
– x 2 = x2 y
Ans.
dx
dy
dx
1 t 3/2
.
+C
2 3/2
2
= x (1 + y)
∴
3C = 126
∴
The particular solution
is 3y + (25 – x2)3/2 = 126
∴
dy
∫ (1 + y )
= x2 dx
=
∫ x dx + c
=
x3
+c
3
∴
when x = 4, 3y + (25 – 16)3/2 = 126
∴
3y = 99 ⇒ y = 33
2
GROUP (C)–HOME WORK PROBLEMS
∴
log (1+ y )
Q-1) Show y = sin ax and y = cos ax are
is the general solution
Now, x = 0, when y = 2
log (1 + 2) =
x3
+ log 3
3
log(1 + y) – log 3 =
...(i)
Differentiating twice w.r.t. x we get
dy
dx
=
d 2y
+ a 2y = 0 .
dx 2
Ans. Consider y = sin ax
03
+c
3
log(3) = c
log(1 + y)
solutions
x3
3
3
1+ y  x
log 
=

 3  3
is the required particular solution
= a cos ax
d 2y
= –a2(sin ax)
dx 2
d 2y
= –a2y
dx 2
... By (i)
d 2y
+ a2y = 0
dx 2
Consider y = cos ax
... (i)
Diff. twice w.r.t. x we get
dy
= x 25 – x 2 when x = 0, when y = 1 3 .
Q-12)
dx
Hence find y, when x = 4
Ans.
dy
dx
2
= x 25 – x
dy
dx
= –a cos ax
d 2y
= – a2 (cos ax)
dx 2
d 2y
= – a2y
dx 2
Differential Equation
Mahesh Tutorials Science
12
d 2y
+ a2y = 0
dx 2
x2
d 2y
dy
+x
= –y ⇒
2
dx
dx
This shows that y = sin ax and y = cos ax
both are solutions of D.E.
x2
d 2y
dy
+x
+y = 0
dx
dx 2
d 2y
+ a2y = 0
dx 2
This shows that y = m sin (log x) + n cos (log x) is
a solution of
x2
Q-2) Verify that y = ae x is a solution of
d 2y
– y = 0 . find particular solution when
dx 2
x = 0, y = 1.
Q-4) Verify that y = ae 2x is a solution of
dy
– y log y = 0 and find the particular
dx
solution when x = 1, y = e
x
x
Ans. y = ae
Diff. w.r.t.x
Ans. y = ae2x
dy
= aex
dx
... (i)
dy
= 2ae2x
dx
Differentiating again w.r.t.x
d 2y
= aex
dx 2
∴
... (ii)
d y
– y = aex – aex
dx 2
... By (i) & (ii)
y
= e 2x
a
=0
= R.H.S.
log(y\a) = 2x
( log y – log a ) = 2
y = aex is solution of give D.E
Now, when x = 0, y = 1 from (i) we get,
1 = ae
a=1
dy
= 2y
dx
y = ae2x
2
L.H.S.
d 2y
dy
+x
+y = 0
2
dx
dx
x
0
dy  log y – log a 
=
 .y
dx 
x

The particular solution is y = ex
x
Q-3) Show that solution of the differential
When x = 1, y = e
e = ae2
2
equation x 2
d y
dy
+x
+ y = 0 is
2
dx
dx
y = m sin (log x) + n cos (log x)
Ans. Consider y = m sin (log x) + n cos (log x) ... (i)
Diff. w.r.t. x
we get,
dy m cos ( log x ) n sin ( log x )
=
–
dx
x
x
x
∴
dy
= m cos(log x) – n sin(log x)
dx
d 2y
dy
+x
= –[m sin(log x) + n cos(log x)]
2
dx
dx
... By (i)
Differential Equation
e
= e–1
e2
a=
1
e
1 2x
e = e 2x–1
e
This is required particular Solution
y=
2
m sin ( log x ) n cos ( log x )
d 2 y dy
–
+
= –
2
x
x
dx
dx
x2
a=
Q-5) Verify that y = x 2 e x is a solution of
Diff. w.r.t. x we get,
x
dy
– log y . y = 0
dx
d 2 y 1  dy 
= 

dx 2 y  dx 
Ans. y = x2ex
dy
dx
= x2ex + 2xex
d 2y
= x2ex + ex(2x) + 2ex + 2xex
dx 2
Mahesh Tutorials Science
13
2
R.H.S.
=
1  dy 
y  dx 
=
1
(x2ex + 2xex )2
x ex
=
∴
y2
 sin 2y 
+y
–
2
 2 
+ x ( – cos x ) –
∫ ( – cos x ) .1dx = C
2
(
x 2e x
2
) (
)
+ 4 x 3 e 2 x + 4x 2 e 2 x
2
x e
2 x
∴
y 2 y sin 2y 1  – cos 2y 
+
– 

2
2
2
2

x
– x cos x + sin x = C
= x e + 4xe + 2xex
=
 sin 2y 
1 (dy )
2 
∫ 
x
∴
d 2y
dx 2
2y2 + 2y sin 2y + cos 2y
– 4x cos x + 4 sin x = C
is the general equation
= LHS.
Hence verified that y = x2ex is the solution of
d 2y 1  dy 
= 

dx 2 y  dx 
2
Putting x = 0 and y = 0 we get C = 1
The particular solution is
2y2 + 2y sin 2y – cos 2y
– 4x cos x + 4 sin x – 1 = 0
Q-6)
y2 –
2
Ans. y –
∴
∴
∴
dy
dy
= x2
dx
dx
Q-8) (1 + y2)tan–1x dx + 2y (1 + x2)dy = 0
–1
Ans. tan x dx + 2y dy = 0
1+ x 2
1+ y 2
dy
dy
= x2
dx
dx
y2 = (x2 + 1)
∫
dy
dx
dt =
∫
∴
1
tan–1 x + = c
y
=
( tan x )
–1
Q-7) 2y cos2y dy + x sin x dx = 0 when x = y = 0
Ans. 2y cos2y dy + x sin x dx = 0
∴
∴
∫ 2y
∫
2
This is the required general solution
Q-9) (sin x + cos x)dy + (cos x – sin x)dx = 0
Ans. Integrate
cos x – sin x
y dy + x sin x dx = C
(1+ cos2y )
+ log |1 + y2| = c
∫ dy + ∫ sin x + cos x dx = 0
Integrating we get
2
=c
2
2
This is the required general solution
∫ 2y cos
2
t2
+ log |1 + y2| = c
2
–1
+c
y
tan–1 x
1
dx
1+ x2
∫ t dt + log 1 + y
dx
= y –2dy + c
x 2 +1
∴
∫
Put t = tan–1x
dx
dy
= 2
2
x +1 y
∫
tan–1 x
2y
dx +
dy = 0
2
1+ x
1 + y2
∫
dy + x sin x dx =C
∫ ydy + ∫ y cos 2 y dy + ∫ x sin x dx =C
y + log|sin x + cos x| = c
This is the required general solution
Q-10) sec2 x tan ydx + sec2y tan xdy = 0
Ans. Divide by (tan y) (tan x) we get
sec2 x
sec 2 y
dx +
dy = 0
tan x
tan y
Integrate
Differential Equation
Mahesh Tutorials Science
14
∫
sec 2 x
dx +
tan x
∫
sec 2 y
dy = 0
tan y
log |tan x| + log |tan y| = log c
Q-13)ex(y + 1) dx – (ex + 1) dy = 0
log |tan x tan y| = log c
tan x tan y = c
Ans.
sec 2 y
ex
dx +
dy = 0
x
e –1
tan2 y
∫e
∫
e
x
x
–1
dx +
∫
1
cos 2 y
×
dy = 0
2
cos y sin2 y
ex
dx +
e –1
x
x
log |e – 1|+
x
log |e – 1|+
∫e
Ans.
=
∫
∫
cos ec 2y dy = 0
Ans. ∴
log |ex – 1|+ (– cot y) = c
∫
∫
Ans. ∴
∫
1
dt = c
t
Ans. ∴
3
(1 – e )
dy
1 – y2
=0
2y dy
2x dx
2
2
∫ (1 + y ) = – ∫ (1 + x )
= log c
tan y = c(1 – ex)3
This is the required general solution
dy
dx
∫ (1 + y ) = ∫ (1 – x )
∴ log |1 + y| = –log|1 – x| + logc
∴ (1 + y) (1 – x) = c
This is the required general solution
–3 log |1 – e x| + log |tan y| = c
x
∫
(Multiplying both sides by 2)
∴ log |(1 + y2)| = –log|(1 + x2)| + log|c|
∴ (1 + y2)(1 + x2) = c
This is the required general solution
–3 log |1 – e x| + log |t| = c
tan y
1– x
+
2
Q-17) (1 – x)dy – (1 + y) dx = 0.
–3 log |1 – e x| +
log
dx
Q-16) (1 + x2) y dy + (1 + y2) x dx = 0
sec 2 y
dy = 0
tan y
Put tan y = t
dt = sec2y
∫
This is the required general solution
3e x
sec 2 y
dx
+
dy = 0
tan y
1 – ex
ex
dx +
1 – ex
( cos y – sin y ) dy
sin–1x + sin–1 y = c.
log |ex – 1|– cot y = c
3
y
dy
1 – y2
+
=0
dx
1 – x2
∫
Ans. Divide by tan y (1 – ex)
∫ –e
( sin x + cos x ) dx
This is the required general solution
Q-15)
Q-12) 3ex tan xdx + (1 – ex)sec2 ydy = 0
x
ex sin x = –ey cos y + c
ex sin x + ey cos y = c
1
dy = 0
sin2 y
This is the required general solution
dy
∫ (y +1)
Q-14) ex (sin x +cos x) dx + ey (cos y – sin y) dy = 0
2
sec y
dy = 0
tan2 y
ex
dx =
e +1
x
∴
log |ex + 1| = log|y + 1|+ log c.
ex + 1 = c(y + 1)
∴
This is the required general solution
This is the required general solution
Q-11) extan2y dx + (ex – 1) sec2y dy = 0
∫
Ans. ∴
Q-18)
dy
– x 2 y When x = 2, y = 4
dx
Ans.
1
dy = x2.dx
y
1
dy – dx = 0
y
1
∫ y dy – ∫ x dx = 0
Differential Equation
2
Mahesh Tutorials Science
log y –
15
x3
=c
3
Q-21)(x2 + 1)
dy
+ 1 = e–y when x = 1= y
dx
Ans. (x2 + 1)
dy
= e–y – 1
dx
when y = 4, x = 2
3
log y –
x
3
= log 4 –
8
3
3
log y – log 4 –
x
8
+ =0
3 3
2
solution
Q-19) (1 – x)dy – (1 + y) dx = 0 when x = 2, y = 4
1
1
dy –
dx = 0
1+ y
1– x
1
log 1 – x
–1
=C
log |1 + 4| + log |–1| = C
log |5|+ log |–1|= C
log |5| + log |–1|= C
log |–5|= C
log |1 – y| + log |1 – x|= C
log |(1 – y)(1 – x)|= log |– 5|
(1 – y) (1 – x) + 5 = 0
This is the required particular general
solution
Q-20)
dy y + 2
+
= 0 when x = 1, y = 2
dx x + 2
1
1
Ans.
dy +
dx = 0
y +2
x +2
∫ (x )
1
2
2
+ (1)
dx +
∫
–e y
dy = C
1 – ey
tan–1 (1) + log |1 – ey| = C
tan–1 x + log |1 – ey| – log |1 – e|–tan–1(1) = 0
tan–1 x + log
1
∫ y + 2 dy + ∫ x + 2 dx = C ,
log |y + 2| + log |x + 2| = log C,
log (y + 2) (x + 2) = log C,
(y + 2) (x + 2) = C
when y = 2 & x = 1
(4) (3) = C
C = 12
(y + 2) (x + 2) = 12
xy + 2(x + y) = 12
xy + 2(x + y) = 8
This is the required particular general
solution
1 – ey
π
– =0
1–e
4
This is the required particular general
solution
Q-22)(x + 1)
dy
– 1 = 2e–y When x = 1, y = 0
dx
Ans. (x + 1)
dy
– (2e–y + 1) = 0
dx
–1
1
dy +
dx = 0
x +1
2e – y + 1
–1
1
dy +
dx = 0
2
x
+1
+1
ey
∫
–e y
dy +
2 + ey
1
∫ x +1 dx = C
–log |2 + ey|+ log |x + 1|= C,
Integrating
1
ey
dy = 0
1 – ey
tan–1 (x) + log |1 – ey| = C
∫ 1 + y dy – ∫ 1 – x dx = C
log |1 + y| –
2
dx –
Integrating
This is the required particular general
1
1
( x ) + (1)
3
y  x – 8
log   =
3
4
Ans.
1
1
dx – – y
dy
x +1
e –1
2
log
x +1
= log C
2 + ey
x +1
=C
2 + ey
1 +1
=C
2 + e0
2
=C
3
2
x +1
=
y
3
2+e
3(x + 1) = 2(2 + ey)
3x + 3 = 4 + 2ey
3x – 1 = 2ey
Differential Equation
Mahesh Tutorials Science
16
Q-23) 3ex tan y dx + (1 + ex) sec2y dy = 0 when
∴
u +1
du
u +3
∴
∫
∴
∫ 1 – u + 3 du + ∫1dx = c
∴
u – 2log |u + 3| + x = c
∴
(3x – 2y) – 2log|3x – 2y + 3| + x = c
∴
4x – 2y – 2log|3x – 2y + 3| = c
π
x = 0, y =
4
2
x
3e
sec y
Ans.
dx +
dy = 0
x
tan y
1+ e
3
e
∫ 1+ e
x
dx +
sec y
∫ tan y dy = C
x
3 log |1 + e | + log |tan y| = log C
log |1 + ex|3 + log |tan y| = log C
|1 + ex|3 tan y = C
u +3
π
=C
4
Ans. x dx + y dy = (x2 + y2)dx
GROUP (D)–CLASS WORK PROBLEMS
Diff. w. r. t. x
2x + 2y
∴
x dx + y dy =
1
du = u dx
2
...(i)
∴
du
u
∴
∫u
∴
log u = 2x + c1
∴
log u = 2x + c1
∴
u = ec1.e2x
∴
x2 + y2 = ce2x
Put 3x – 2y = u
∴
∴
1
du 
3–
2 
dx 
=
(i) becomes,
1
du 
3–
=

2
dx 
du
=
dx
2u + 3
u +1
4u + 6
u +1
∴
3–
∴
du
4u + 6
= 3–
dx
u +1
=
=
3u + 3 – 4u – 6
u +1
–u – 3
u +1
Differential Equation
1
du
2
Equation (i) becomes
3x – 2y = u
dy
dx
dy
dy
=
dx
dx
∴
dy 6x – 4y + 3
=
by su bs titutin g
dx
3x – 2y +1
2dy du
=
dx
dx
...(i)
Put x2 + y2 = u
solution
Then 3 –

2
substituting x2 + y2 = u
(1 + ex)3 tan y = 8
This is the required particular general
2 ( 3x – 2y ) + 3
Ans. dy = 6x – 4y + 3 =
dx 3x – 2y +1
( 3x – 2y ) +1
∫
du + dx = c
Q-2) Solve x dx + y dy = (x2 + y2)dx by using
C=8
Q-1) Solve
– dx
This is the required general solution
π
when x = 0, y =
4
(1 + 1)3 tan
(u + 3) – 2

2
x
=
du
= 2dx
∫
= 2 dx + c1
where c = ec1
This is the required general solution
dy 

Q-3) Solve  y + x
cos (xy) = sin2x
dx 

substitution xy = u
dy 
Ans.  y + x
cos (xy) = sin2x
dx 

Put xy = u ⇒ y + x
dy
=
dx
...(i)
du
dx
by
th e
Mahesh Tutorials Science
17
Equation (i) becomes
Q-5) Solve
du
cos u = sin2 x
dx
∴
1 – cos 2x
du
cos u =
2
dx
∴
2 cos u du = (1 – cos 2x)dx
∴
∫
2 cos u du =
x+y=u
Ans.
...(i)
Put x + y = u
…(ii)
1
1+
sin 2x
+ C1
2
2 sin u = x –
∴
4 sin xy = 2x – sin 2x + 2C1
∴
4 sin xy = 2x – sin 2x + C
dy du
=
dx dx
dy du
=
–1
dx dx
…(iii)
du
– 1 = sin u + cos u – 1
dx
where 2C1 = C
du
= sin u + cos u
dx
is the required general solution.
 dy
 y
Q-4) Solve  x
− y  e x = x 2 cos x using
 dx

1
du = dx
sin u + cos u +1
y
=u
x
1
∫ sin u + cos u +1 du – ∫ 1dx = c
dy
 y
Ans.  x
– y  e x = x 2 cos x
 dx

y
=u
x
...(i)

 u 
Put t = tan   
 2 

...(ii)
∫
y = ux
dy
du
=u+x
dx
dx
2dt
– x =c
2
+1 + t 2
1
∫ t +1 dt – x = c
log |t + 1| – x = c

 u
2 du
2
ux + x dx – ux  e = x cos x


du u
e
dx
1
 2dt 
.
– x =c
2
2 
2t
1– t
1+ t 
+
+1
1+ t2 1+ t2
∫ 2t +1 – t
 

du 
– ux  eu = x2 cos x
x  u + x

dx 
 

x2
dy
= sin (x + y) + cos (x + y)
dx
Diff w.r.t. x
∫ (1 – cos 2x ) dx + C
∴
dy
= sin (x + y) + cos (x + y) using
dx
 x +y 
log tan 
 +1 – x = c
 2 
= x2 cos x
This is the required general solution.
u
e du = cos x dx
Q-6) (x2 + y2)dx – 2xy dy = 0
∫e
∫
u du = cos x dx
∴
eu = sin x + c
∴
e
y
x
– sin x = c
2
2
Ans. dy = x + y
dx
2xy
Put
y = vx.
...(i)
The above equation is. Homogenous DE
This is the required general solution.
∴
dv
dy
=v+x
dx
dx
∴
the equation (i) becomes
Differential Equation
Mahesh Tutorials Science
18
∴
v +x
dv
dx
=
x 2 + v 2x 2
2xvx
v +x
dv
dx
=
1+ v 2
2v
=
dx
x
2vdv
1 – v2
sec2 u
du =
tan u
∫
∴
log (tan u) = – log x + log C
∴
log (x tan u) = log C
∴
x tan u = C
∴
x tan
Integrating, we have
∴
∫
2vdv
1 – v2
=
∫
y
=C
x
Ans. (x3 + y3)dx – 3xy2 dy = 0
–log c
x 3 + y3
dy
=
3xy 2
dx
2
x (1 – v )|= c
  y 2 
x 1 –    = c
  x  
dy
dx
∴
equation (i) becomes
x2 – y2 = cx
= v+x
v+x
(
)
(
)
3x x 2v 2
x3 1+v3
=
x
∴
3x 3v 2
tan
=
1 + v 3 – 3v 3
3v 2
=
1 – 2v 3
3v 2
...(i)
∴
3v 2
dv =
1 – 2v 3
dx
x
dx
x
Put y = ux
dy
dx
du
= u+x
dx
∴
3v 2
dv =
1 – 2v 3
u+x
du
dx
∴
dx
3v 2
–
dv = 0
x 1 – 2v 3
∴
–6v 2
dx 1
+ .
=0
x
2 1 – 2v 3
du
dx
∴
x
∴
sec2 u
du
tan u
tan u
= u–
sec 2 u
=
–
tan u
sec2 u
=
–
dx
x
3
=
3v 2
(
)
Integrating
∴
Differential Equation
(1 + v )
dv
1+ v 3
=
–v
dx
3v 2
The above equation is Homogenous DE
y
dy y
x
= –
dx x
2 y
sec
x
x 3 + x 3v 3
dv
=
dx
y
y
y

– y sec2  dx + x sec2
dy = 0
Q-7)  x tan
x
x
x

y
y
y sec 2 – x tan
dy
x
x
Ans.
=
dx
x sec2 y x
dv
dx
∴
is the required general solution
∴
...(i)
Put y = vx
x2 – y2
=c
x
∴
∫x
Q-8) (x3 + y3)dx – 3xy2 dy = 0
– log |1 – v2| = log|x| + log c
∴
–
This is the required general solution.
dx
log c
x
log |x (1 – v2)|=
dx
∴
∫
dx 1
+
x
2
∫
d
1 – 2v 3
dv
dv = logc1
1 – 2v 3
(
)
Mahesh Tutorials Science
∴
19
1
log(1 – 2v3) = log c1
2
log x +
3
u=x–y
= log c1
... (ii)
du
dx
=
dx dy
–
dx dx
=
1–
∴
log x 1 – 2v
∴
x . 1 – 2v 3 = c1
du
dx
∴
3

y  
x 2 1 – 2    = c12
 x  

 dy 
–
 =
 dx 
∴

2y 3 
x 1 – 3  = C
x 

2
2
1
...(where C = c )
dy
dx
dy
dx
du
–1
dx
–dy
+1
dx
=
Put (ii), (iii) in (i)
∴
 x 3 – 2y 3 
x2 
 =C
x3


 –du

u2 
+1
dx


∴
x3 – 2y3 = Cx
–du
+1 =
dx
This is required general solution.
GROUP (D)–HOME WORK PROBLEMS
dy
= x 2 + y 2 using x2 + y2 = u
dx
Q-1) Solve x + y
(given)
du
dx
dy
2x + 2y
=
dx
dy
x +y
dx
∫
∫
a2 – u2
dy
=
dx
u2
∫
∫
1 du
=u
2 dx
1
du –
2u
a2
u2
a2
–1
u2
dy
=
dx
∫
1 du
2 dx
=
= a2
u2
du = dx
u2 – a2
dy
Ans. x + y
= x2 + y2
dx
x2 + y2 = u
...(iii)
u2
du = dx
u2 – a2
(u
2
)
– a2 + a2
2
u – a2
du +
∫ (u
du = x + C
a2
2
– a2
)
du = x + C
 1
u –a 
u + a2 
log
 = x +C
2
a
u +a 

dx = 0
 1
u –a 
x – y + a2 
log
 = x +C
2
a
u +a 

1
log|u| – x = C1
2
1
log|x2 + y2| = C1 + x
2
 1
x –y –a
x – y + a2 
log
 2a
x – y +a

log |x2 + y2| = 2C1 + 2x
2c
x2 + y2 = e 1 + 2x
x2 + y2 = e2x . C
x–y –

 = C

a2
x –y –a
log
=C
2a
x – y +a
2c
where c = e 1
This is the required general solution
Q-2) Solve ( x – y )
Ans.
2
(x – y )
2
dy
= a 2 using x – y = u
dx
dy
= a2
dx
2x – 2y = a log
x –y –a
+C
x – y +a
This is the required general solution
... (i)
Differential Equation
Mahesh Tutorials Science
20
Q-3) Solve x dx + y dy = (x2 + y2) dx by using
x2 + y2 = u
1
du – dx = 0
1 + cos u
Ans. x dx + y dy = (x2 + y2). dx.
x +y
dy
dx
1
du = dx
1 + cos u
= x2 + y2
...(i)
Put, x2 + y2 = u
1
∫ 1 + cos udu – ∫1dx = C
1
dy
dx
∴
x +y
∴
1 du
=u
2 dx
∴
∴
∴
∴
∴
1 du
2 dx
=
x+C
1
u
sec2
=
2
2
x+C
1
2
tan
1
tan
1
log |u| = x + c
2
u
2
=
u
2
2
=x+C
This is the required general solution
2
log |x + y |= 2x + c
e
2x + c
2
Q-5) Solve (x + y)2
2
=x +y
∴
e2x.c = x2 + y2
2
x + y = c .e
dy
= 1 by substitution
dx
x+y=t
2x
2
x+C
=x+C
tan ( x + y )
1
log |x2 + y2| = x + c
2
2
u
2
1
2
∫ 2u du = ∫ dx
e .ec = x2 + y2
Ans. (x + y)2
2x
This is the required general solution
dy
Q-4) Solve
= cos(x + y) by the substitution
dx
dy
=1
dx
dx dy dt
+
=
dx dx dx
dy dt
=
dx dx
dy
= cos(x + y)
dx
... (i)
dy dt
=
–1
dx dx
x+y=u
... (ii)
Put (ii) (iii) in (i)
dx dy
+
dx dx
du
dx
=
du
dx
= 1+
dy
dx
du
dx
= 1+
dy
dx
Put (ii), (iii) in (i)
du
– 1 = cos u
dx
du
= cos u + 1
dx
Differential Equation
(x + y )
2
 dt

 dx – 1 = 1


 dt

t2 
– 1 = 1
 dx

... (iii)
... (iii)
... (i)
Put x + y = t
1+
x+y=u
Ans.
=
2
∫
{Substituting in (i)}
∴
∴
∫ 2 cos
du
dx
dy
2x + 2y
=
dx
dt
1
–1= 2
dx
t
dt
1
=
+1
dx t 2
dt 1+ t 2
= 2
dx
t
... (iii)
Mahesh Tutorials Science
21
t2
dt = dx
1+ t 2
Q-7) S olve
Ans.
Integrate
∫
t2
dt – 1dx = C
1+ t 2
∫
1
∫
1+
dt – x = C
2
y – tan–1 (x + y) – x = C
This is the required general solution
dy 

Q-6) Solve  y + x
cos(xy) = sin 2 x by the
dx 

substitution xy = u
Diff. x
dy
du
+y =
dx
dx
du
dx
=
du
–1
dx
... (iii)
du
dx
=
u +1 + u – 1
u –1
du
dx
=
2u
u –1
u –1
du = dx
2u
u –1
du – dx = 0
2u
... (ii)
∫ 2 – 2u du – ∫1dx = 0
... (iii)
1
1
u – log|u|–x = C
1
2
2
∫
1
1
u – log |u| – 2x = 2C1
x + y – log |x + y| – 2x = C ; where C = 2C1
cosu du = sin2x dx
∫
du
dx
∫
du
cos u = sin2x
dx
2
... (ii)
... (i)
Put (ii) & (iii) in (i)
∫ cos u du – ∫ sin
... (i)
u +1
du
–1 =
u –1
dx
t – tan–1 (t) – x = C
x + y – tan–1 (t) – x = C
Put xy = u
dy
dx
dy
=
dx
1
1–
dt – x = C
1+ t2
dy 

Ans. y + x
cos (xy) = sin2x
dx 

x + y +1
x + y +1
=
dx dy
+
=
dx dx
1+ t – 1
dt – 1dx = C
1+ t2
∫1 – 1 + t
dy
dx
x+y = u
∫
2
by s u bstitu tio n
x+y=u
t2
dt – dx = 0
1+ t 2
∫
dy x + y +1
=
dx x + y – 1
y – x – log |x + y| = C
This is the required general solution
xdx = 0
1 – cos 2x
cos u du –
dx = C1
2
∫
Q-8) xy2 dy – (y3 – 2x3) dx = 0
Ans. xy2 dy – (y3 – 2x3) dx = 0
dx
– (y3 – 2x3) = 0
dy
sin u –
1
1
1dx +
cos 2xdx = C1
2
2
xy2
1
1 sin 2x
x+
= C1
2
2 2
It is homogeneous DE in 3rd degree
sin u –
∫
sin ( xy ) –
∫
x sin 2x
+
= C1
2
4
... (i)
Put y = vx
... (ii)
dy
dv
=v + x
dx
dx
... (iii)
4 sin (xy) – 2x + sin 2x = 4C1
Put (ii) & (iii) in (i)
4 sin (xy) – 2x + sin 2x = C,
where C = 4C1
This is the required general solution
dy 

x(vx)2 v + x
– (vx)3 – 2x3 = 0
dx 

v 3x 3 + v 2x 4
dy
– v3x3 + 2x3 = 0
dx
Differential Equation
Mahesh Tutorials Science
22
dy
+ 2x3 = 0
dx
v 2x 4
v2
y2  y

+1 x 4 = C
2 
x

x 
2
dy
+
=0
x
dx
xy2 (x + y) = C
This is the required general solution
–2
dx
x
v2 dv =
∫
u2 (u + 1) x4 = C
∫
v3
+ 2log |x| = C
3
y 
x
 
3
2
Ans. x
Put y = vx
+ 2log |x| = C
dy
dx
3
y
+ 2log |x| = C
3x 3
This is the required general solution
Ans. (y2 + 2xy) dx + (2x2 + 3xy) dy = 0
 y 2 + 2xy 
dy
= – 2

dx
 2x + 3xy 
... (i)
Put y = u x
dy
dy
=u+x
dx
dx
the equation (i) becomes
u+x
dy
dx
=–
=
–
u 2 x 2 + 2x 2u
2x 2 + 3x 2u
= v+x
dv
dx
∴
dv 

x 2 v + x
= x (vx) + v2x2
dx 

∴
dv 

= x2 (v2 + v)
x 2 v + x
dx 

∴
x
∴
∫v =∫ x
∴
–1
= log|x| + c
v
∴
–x
= log|x| + c
y
Q-9) (y2 + 2xy) dx + (2x2 + 3xy) dy = 0
∴
dy
= xy + y2
dx
This is a homogeneous Differential
equation
3
Diff.
dy
= xy + y 2
dx
2
Q-10) x
1
v dv + 2 dx = 0
x
2
dv
= v
dx
dv
2
dx
2
This is the required general solution
u 2 + 2u
2 + 3u
du
x
dx
u 2 + 2u
–u
= –
2 + 3u
du
x
dx
–4u 2 – 4u
=
2 + 3u
 dy

– y  .e y /x = x
Q-11)  x
 dx

y
∴
dx
2 + 3u
du = –4
x
u ( u +1)
∴
∫ u (u +1) du
2 + 3u
2
=
1 
–4
 dy

– y e x = x
Ans.  x
 dx

y = vx
dx
∫x
 

dv 
– (vx )  e v = x
 x v + x

dx 
 

dx
∫  u + u +1  du – 4∫ x
... by Partial Fraction
∴
dv 
 dy
 dx = v + x dx 


2 log u + log (u + 1) = –4log x + log C
logu2 + log (u + 1) + log x4 = log C
∴

 v
2 dv
 xv + x dx – vx  e = x


∴
x2
∴
∫ e dv = ∫ x
log [u2 (u + 1)x4] = log C
Differential Equation
dv v
e =x
dx
v
dx
Mahesh Tutorials Science
∴
∴
ev
23
= log |x| + c
y/x
e = log |x| + c.
This is the required general solution
Q-12) x sin
Ans. ∴
dy
dx
y
y

dy +  x – y sin
x
x

∫
= e
Pdx
1
= e

 dx = 0

∫ cotx + x dx
= elog(sin x) + log x
= elog(sin x(x))
I.F
y 
– x + y sin  
x
=
y 
x sin  
x
= x sin x
y I.F =
∫ Q (I.F ) dx
1
∫x
y(x sin x) =
y = vx
dy
dv
=v + x
dx
dx
x sin x dx
xy sin x = – cos x + c
xy sin x + cos x = c
This is a homogeneous Differential
equation
Put y = v x
This is the required general solution
Q-2) (x + y)
dy
dv
=v + x
dx
dx
∴
x
dv
+v =
dx
– x + vx sin v
x sin v
∴
x
dv
+v =
dx
–1 + v sin v
sin v
∴
x
dv
+v =
dx
–1
+v
sin v
∴
∫ sinv dv = ∫ x
∴
cos v = log |x| + c
∴
cos
Ans. (x + y)
dx
dy
P = –1 ; Q = y
GROUP (E)–CLASS WORK PROBLEMS
dy
Q-1) x sin x
+ (x cos x + sin x)y = sin x
dx
( x cos x + sin x ) =
Ans. dy +
y
dx
x sin x
dy
+ Py = Q
dx
= x+y
dx
+ Px = Q
dy
This is the required general solution
Compare with,
=
sin x
x sin x
1
x
I.F.
∫
= e
Pdy
∫
= e
–1dx
= e–y
Soln is
x I.F =
∫ Q I.Fdy
x e–y =
∫ ye
–y
dy
∫
–y
x e–y = y e dy –
x e–y =
1
1
;Q=
x
x
dx
dy
=
Compare with
y
= log |x| + c
x
dy 
1
+  cot x +  y
dx 
x
dy
=1
dx
dx
– 1x = y
dy
dx
P = cot x +
I.F.
ye – y
–
–1
∫
 dy
∫  dy ∫ e
–y

dy  dy

1e – y
dy
–1
∫
–y
x e–y = – y e–y + e dy
Differential Equation
Mahesh Tutorials Science
24
x e–y = – y e–y +
x
= –y –1+
dx
+ Px = Q
dy
e –y
+c
1
c
e –y
= c ey
x+ y+1
tan –1 y
1
;
Q
=
1+ y2
1 + y2
P=
= e∫
I.F.
This is the required general solution
Pdy
1
Q-3)
= e
2
y dx + (x – y )dy = 0
Ans. y dx + (y2 – x)dy
tan
= e
y2 – x
y
dx
dy
=
dx
dy
x
= y–
y
∫ 1+y 2 dy
∴
–1y
solution is
x I.F
xe
tan –1y
=
∫ Q I.Fdy
=
∫
1
dx
x = y
+
y
dy
Compare with
e tan
–1y
tan –1 y
dy
1 + y2
tan–1 y = t
1
dy = dt
1 + y2
dx
+ Px = Q
dy
∫e
P =
1
; Q=y
y
x e tan
–1y
=
I.F.
= e∫
x e tan
–1y
t
= t e dt –
x e tan
–1y
t
t
= t e – 1e dt
x e tan
–1y
= t et – et + c
x e tan
–1y
–1
tan
= tan y e
x
= tan–1y –1 +
P dy
t
t dt
∫
 dt

∫  dt ∫ e dt 
t
1
= e
∫ y dy
∫
= elog y
= y
Soln is
x I.F =
∫ Q I.Fdy
xy
=
∫
xy
=
∫
y 2dy
=
y3
+c
3
tan –1y
tangent to the curve at each of its point is
equal to the sum of the abscissa and the
product of the abscissa and ordinate of the
point.
Ans. Since
dy tan–1 y – x
+
dx
1 + y2 1 + y2
dy
represents the slope of tangent to
dx
a given curve at a point (x ,y) the given
equation is
tan –1 y
1 + y2
Compare
∴
Differential Equation
+c
Q-5) Find the equation of the curve passing
through the point (0,1), if the slope of the
Q-4) (1 + y2)dx = (tan–1y – x)dy
dy
1
+
x =
dx 1 + y 2
–1y
This is the required general solution
This is the required general solution
Ans.
– e tan
c
e
y y dy
xy
–1y
dy
dx
= x + x.y
dy
dx
= x(1 + y)
Mahesh Tutorials Science
∴
25
dy
= xdx
1+ y
– ( x – 1) e –3 x
=
3
Integrating both sides, we get
∫
∴
dy
1+ y
=
∫
∴
log |1 + y| =
∴
1 + y = ex
2
∴
1 + y = ex
2
x2
+ c1
2
c
2
+e
2
+ ec
1
∴
Equation becomes, 1 + y = e
∴
y= e
x2
2
0=
1 1
– +c
3 9
∴
c=
2
9
∴
Equation becomes
i.e. ye
+2
Q-1)
– ( x – 1) e – x
–3x
3
( –e ) – 2
+
–3 x
9
–x
= –3(x –1)e –e
–3x
9
–2
dy
+ y = e –x
dx
Ans. Compare with
dx
+ Py = Q
dy
dy
represents the slope of tangent to
dx
P = 1 ; Q = e–x
∴
equation is
I.F.
= e∫
Pdx
= e ∫ 1dx
= x + 3y – 1
= ex
solution is
dy
– 3y = x – 1
dx
y (I.F)
=
∫ Q I.Fdx
dy
+ Py = Q
dx
yex
=
∫e
Where, P = – 3, Q = (x – 1)
yex
=
∫ 1dx
yex
= x+c
y
=
The given equation is of the form
= e ∫ Pdx = e –3 ∫ d x
∴
I.F.
∴
Solution is
∴
–e –3 x
+c
9
+1
a given curve at a point (x ,y) the given
∴
+
GROUP (E)–HOME WORK PROBLEMS
2
passes through the origin and has the slope
x + 3y – 1 at any point (x, y) on it.
dy
dx
3
∴
ye–3x =
Q-6) Find the equation of the curve which
Ans. Since
=
∫
Curve passes through (0,0),
ec = 2
x2
– ( x – 1) e – x
x dx
But curve passes through (0 ,1)
∴
ye
–3x
1
e –3 x dx
3
+
= e–3x
∴
y(I.F.)
=
∫ Q (I.F.) dx
ye–3x
=
∫ ( x – 1) e
=
( x – 1) ∫ e –3x dx –
∫
–3 x
–x
e x dx
x
+ ce–x
ex
This is the required general solution
dx
Q-2)
dy
+ y sec x = tan x
dx
Ans. Compare with
 d

–3 x
 dx ( x – 1) e dx 


∫
dx
+ Py = Q
dy
P = sec x ; Q = tan x
Differential Equation
Mahesh Tutorials Science
26
∴
I.F.
= e ∫ Pdx
∫
 d

– 
∫  dx log x ∫ x dx  dx
3
= log x x dx
yx2
= e ∫ secx dx
= e
3
log (sec x + tan x)
= sec x + tan x
y (I.F) =
∫ Q I.Fdx
y(sec x + tan x) =
∫
sec x ( sec x + tan x ) dx
y(sec x + tan x) =
∫
sec 2 x + sec x tan x dx
yx2
=
( log x )
x4
–
4
yx2
=
( log x )
x4 1
–
4
4
yx2
=
( log x )
x4 1 x4
–
+c
4
4 4
yx2
=
y(sec x + tan x) = tan x + sec x + c
sec x + tan x
c
+
sec x + tan x sec x + tan x
y =
y = 1+
c
sec x + tan x
Ans.
dy
+ 2y = x2 log x
dx
Q-4) ( x + a )
Ans.
=
dy 2
+ y
dx x
= x log x
Compare with
P=
= e∫
∴
= e ∫ Pdx
I.F.
–3
= e ∫ x+a
2
= e log ( x+a )
dx
y (I.F) =
y
log x 2
(I.F) = x2
∫ Q (I.F ) dx
1
(x + a )
y
∫
–3
= (x + a)–3
= e2 log x
= e
dx
= e–3 log (x + a)
Pdx
= e∫ x
5
(x + a )
(x + a )
=
–3
; Q = (x + a)4
x +a
2
P=
; Q = x log x
x
I.F.
x4
+c
16
dy
– 3y = (x + a)5
dx
dy
3
–
y
dx x + a
dx
+ Py = Q
dy
∴
3
dx
+ Py = Q
dy
x 2 log x
x
dy 2
+ y
dx x
4
–
∫ x dx
This is the required general solution
This is the required general solution
Q-3) x
( log x ) x 4
∫
1 x4
dx
x 4
Q ( I.F ) dx
y (I.F)
=
yx2
=
∫
yx2
=
∫ x log x dx
x log x 2dx
(x + a )
y
(x + a )
3
3
3
3
1
=
∫ (x + a )
=
∫ ( x + a ) dx
=
(x + a )
2
4
(x + a )
3
dx
2
+c
2y = (x + a)5 + 2c (x + a)3
This is the required general solution
Differential Equation
Mahesh Tutorials Science
27
Q-5) dr + (2r cot θ + sin 2θ) dθ = 0
dy
2x
+
y =
dx 1 – x 2
dr
+ 2r cot θ = – sin 2θ
dθ
Ans.
(
x 1 – x2
)
–1
2
dy
+ Py = Q
dx
dr
+ 2(cot θ)r = – sin 2θ
dθ
P=
Compare with
dr
+ Pr = Q
dθ
∴
2x
2
; Q = x 1– x
1 – x2
I.F.
(
= e∫
)
–1
2
Pdx
2x
P = 2 cot θ ; Q = – sin 2θ
∴
I.F.
= e∫
Pd θ
= e∫
2 cot θd θ
= e
–2 x
= e – ∫ 1– x 2 dx
= e –log (1 – x
= e2 log sinθθ
log sin
= e
2θ
Soln is
r (I.F)
=
∫ Q (I.F ) θ
r sin2θ
=
∫ – sin 2 θ sin
r sin2θ
=
∫
2
r sin θ
sinθ =
Put
∴
=
cos θ dθ
∫
–2 sin θ cos θd θ
t
= dt
∫
r sin2θ
=
–2
(
r
=
=
4
=
x 1 – x2
)
=
1
1 – x2
(
(
) dy
dx
y (I.F)
=
∫ Q (I.F ) dx
y
1
=
1 – x2
∫ (
)
y
1
=
1 – x2
∫ (
)
∴ x dx =
) +c
y
1 – x2
x 1 – x2
x 1 – x2
–1
2
–3
2
1
dx
1 – x2
dx
=
(
+ 2xy = x 1 – x 2
(
)
(1 – x )
x 1 – x2
2
1
2
–dt
2
∫
t
–3
2
–
dt
2
–1
sin2 θ
c
+
4
sin2 θ
dy
2x
+
y =
Ans.
dx 1 – x 2
–1
Solution is
y
1 – x2
=
y
1 – x2
=
This is the required general solution
Q-6) 1 – x 2
–1
)
log 1 – x 2
∴ –2x dx = dt
t4
+c
4
– sin4 θ
∴
)
Put 1 – x2 = t
3
–2 t dt
r sin θ
θd θ
3
=
2
2
–2sin θ cos θ sin2 θd θ
r sin θ
2
I.F
2
(
= e
= sin2θ
∴
∫ 1– x 2 dx
)
1
2
–1 t 2
+c
2 –1
2
+1
1 – x2
+c
(
2
2
y = + 1– x +c 1– x
)
This is the required general solution
Differential Equation
Mahesh Tutorials Science
28
Q-7) Find the equation of a curve passing
through the point (0,2), given that the sum
of the co–ordinates of any point on the
curve exceeds the magnitude of the slope
of the tangent to the curve at that point
by 5.
dy
represents the slope of tangent to
Ans. Since
dx
a given curve at a point (x ,y) the given
equation is
dy
+5
dx
∴
= x+y
is the general solution since the curve is
passing through the point (0,2).
∴
x=0 y=2
∴
0 + 2 – 4 = c.e0
∴
c = –2
∴
x + y – 4 = –2e–x
∴
y = 4 – x – 2e–x
is the required equation of the curve.
Q-8) The slope of the tangent to the curve at
any point is equal to y + 2x. Find the
equation of the curve passing through the
origin.
dy
– y = (x – 5)
dx
Ans. Since
The given equation is of the form
dy
represents the slope of tangent to
dx
a given curve at a point (x ,y) the given
∴
∴
dy
+ Py = Q
dx
equation is
Where, P = – 1, Q = (x – 5)
dy
dx
= e ∫ Pdx = e –1 ∫ d x
I.F.
∴
Solution is
y(I.F.)
∴
= e–x
ye–x
=
∫ Q (I.F.) dx
=
∫ ( x – 5) e
–x
dy
+ Py = Q
dx
dx
Where, P = – 1, Q = 2x
∫ ( x – 5 ) .e
–x
∫
dx =
x .e
–x
∫
–x
dx – 5 e dx
dx – 5
∫e
–x
I.F.
∴
Solution is
= x.
=
=
– x .e – x –
– x .e
–x
∫
ye–x
∫ 2x.e
∫
dx + 5e
=
–x
= e–x
dx
∫
–x
= 2x . e –
e –x
dx + 5e – x
–1
+ e
–1 dx
∫ Q (I.F.) dx
–x
∫
–x
= e∫
y(I.F.) =
e
e
 d

– 
x e – x dx  . dx – 5
–1
–1
 dx

∫
Pdx
dx
∴
–x
∫
= e
∴
Integrating both sides, we get
∫ x.e
dy
– y = 2x
dx
The given equation is of the form
Consider
–x
= y + 2x
=
–x
 d
∫  dx 2x.∫ e
∫
–2x .e – x + 2 e – x dx
= –2x.e–x – 2e–x + c
= –x.e–x – e–x + 5e–x + c
∴
= –x.e–x + 4e–x + c
∴
y.e–x = –2(x + 1)e–x + c
y = 2(x + 1) + ce–x
Curve passes through (0,0).
∴
y.e–x = –x.e–x + 4e–x + c
∴
y = –x + 4 + ce–x
∴
0 = – 2(0 + 1) + ce0
x + y – 4 = ce–x
∴
c=2
∴
Equation of curve becomes
Differential Equation
–x

dx  . dx

Mahesh Tutorials Science
y = – 2(x + 1) + 2ex
∴
29
Q-2) If a body cools from 800 C to 500 C at room
temperature of 250 C in 30 minutes, find
the temperature of the body after 1 hour.
y + 2(x + 1) = 2ex
GROUP (F)–CLASS WORK PROBLEMS
Ans. Let ‘x’ be the temperature of the body at time t.
Q-1) In a certain culture of bacteria the rate of
increase is proportional to the number
present. If it is found that the number
doubles in 4 hours, find the number of
times the bacteria are increased in 12
∴
dx
∝ (x – 25)
dt
∴
dx
dt
hours.
∴
∴
=
dN
dt
= k.N
∴
log|80 – 25| = k(0) + c
∴
c = log 55
dN
N
= kdt
log|x – 25| = kt + c
When t = 0, x = 80
dN
= k . 1dt
N
∫
When t = 30, x = 50
∴
log|50 – 25| = 30k + log 55
∴
30k = log 25 – log 55
∴
k=
1
 5 
log  
30
 11 
log N = kt + c
When t = 60
When t = 0, N = N0
∴
log N0 = k(0) + c
∴
c = log N0
∴
 1 
 5 
log   + log 55
log|x – 25| = 60 

 30 
 11 
 25 
= log 
 + log 55
 121 
When t = 4, N = 2N0
∴
log (2N0) = 4k + log 2N0
∴
log (2N0) – log N0 = 4k
∴
∴
= log
 2N0 
log 
 = 4k
 N0 
k=
1
log 2
4
When t =12
∴
log N0
x – 25
∴
x = 25 + 11.36
∴
x = 36.360C
12
log 2 + logN0
4
Thus temperature of body after 1 hr. is
36.360C
Q-3) The rate of disintegration of a radio active
element at any time t is proportional to
its mass at that time. Find the time during
= log 8 + logN0
∴
log N0
N
= log(8N0)
which the original mass of 1.5 gm. will
disintegrate into its mass of 0.5 gm.
= 8N0
Thus, in 12 hour, bacteria increases to 8
times.
=
25
× 55
121
125
21
∴
= 12k + log N0
=
∴
= k dt
∫ 1dt
∴
Integrating both sides, we get
∴
dx
∫ x – 25
dN
∝N
dt
∫
dx
x – 25
Integrating both sides, we get
Ans. Let N be the number of bacteria present at
time t.
∴
∴
= k(x – 25)
Ans. Let M be the mass present at time ‘t’.
∴
dM
∝M
dt
Differential Equation
Mahesh Tutorials Science
30
∴
dM
dt
The amount left after one more hour is 16/3
= – kM
gms.
(–ve sign indicates disintegration)
∴
dM
M
Q-5) A body cools according to Newton’s law
from 1000c to 600c in 20 minutes. The
= – kdt
temperature of the surrounding being 200C
how long will it take to cool down to 300C?
Integrating both sides, we get
∫
∴
Ans. Let ‘x’ be the temperature of the body at time
dM
= –k dt
M
∫
t.
log M = –kt + c
∴
dx
∝ (x – 20)
dt
∴
dx
dt
∴
dx
x – 20
When t = 0, M = 1.5
∴
log 1.5 = –k(0) + c
∴ c = log 1.5
When M = 0.5,
∴
log (0.5) = –kt + log (1.5)
∴
 1.5 
kt = log (1.5) – log (0.5) = 

 0.5 
∴
1
t=
log 3
k
present at that instant. Initially, there are
27 gms of certain substance and three
hours later is found that 8 gms are left.
Find the amount left after one more hour...
Ans. Let ‘x’ be the mass present at time ‘t’.
∴
dx
∝x
dt
∴
dx
dt
= k dt
Integrating both sides, we get
dx
∫ x – 20
∴
Q-4) The rate of decay of certain substance is
directly proportional to the amount
= k(x – 20)
∫
= k dt
log|x – 20| = kt + c
When t = 0, x = 1000C
∴
log|100 – 20| = k(0) + c
∴
c = log 80
When t = 20, x = 60
∴
log|60 – 20| = k (20) + log 80
∴
20k = log 40 – log 80
 40 
= log 

 80 
= – kx
1
1
log  
20
2
∴
k=
∴
log|x – 20| =
(–ve sign indicates decay)
When t = 4,
∴
log x =
3
–4 log   + log 27
2
 16 
= log 
 + log 27
 81 
 16

× 27 
= log 
81


 16 
= log 

 3 
∴
x
=
16
gms
3
Differential Equation
1
1
log   t + log 80
20
2
where x = 300 , t = ?
1
1
log   t + log 80
20
2
∴
log|30 – 20| =
∴
log 10 =
∴
log10 – log 80
1
1
t =
log  
20
2
1
1
log   t + log 80
20
2
20 (1 – log 80 )
=
– log ( 2)
Mahesh Tutorials Science
t =
31
Q-7) An equation relating to stability of an
20 ( log 80 – 1)
log 2
aeroplane is
20 ( log 80 – 1)
it will take
log 2
the velocity g, α, k being constants. Find
a expression for the velocity if v = 0 when
minutes to cool
down to 300.
t = 0.
Q-6) Ass ume th at a s ph erical raind rop
evaporates at a rate proportional to its
surface area. If its radius originally is 3mm
and 1 hour later has been reduced to 2mm,
find an expression for the radius of the
raindrop at any time t.
Ans.
dv
= g cos α – kv
dt
∴
dv
+ kv = g cos α
dt
The given equation is of the from
dv
+ Pv = Q
dt
Ans. Let ‘r’ be the radius of the sphere at time t.
4 3
πr , A = 4 πr 2
3
υ =
dv
= g cos α – kv, where v is
dt
Where P = k, Q = g cos α
dv
∝A
dt
= e ∫ Pdt = e ∫ kdt = ekt
∴
I.F.
∴
Solution is
∴
dv
dt
= k.A
∴
v(I.F.)=
∫ Q (I.F ) dt
∴
dv
dt
= k.(4πr2)
∴
vekt
=
∫ g cos α e
∴
vekt
= g cos α
=
4 3
πr
3
dv
dt
=
4
dr
π(3r2).
3
dt
k(4πr2)
= 4 πr 2
But, υ
∴
∴
∴
dr
dt
Integrating both sides, we get
∴
0
=
g cos α
+c
k
∴
c
=
– g cos α
k
∴
vekt
=
g cos α kt g cos α
.e –
k
k
∴
v
=
g cos α g cos α kt
–
.e
k
k
∴
v
=
g cos α
1 – e kt
k
∫
... (i)
When t = 0, r = 3
∴
3 = k(0) + c
∴
c=3
e kt
+c
k
∴
= k dt
r = kt + c
dt
When v = 0, t = 0
dr = kdt
∫ dr
kt
(
)
GROUP (F)–HOME WORK PROBLEMS
When t = 1, r = 2
∴
2 = k(1) + 3
∴
k = –1
∴
Equation (i) becomes r = –t + 3
i.e. r = 3 – t (0 < 1 < 3)
Q-1) If the population of a country doubles in
60 years, in how many years will it be triple
under the assumption that the rate
increase is proportional to the number of
inhabitants?
Given : log 2 = 0.6912, log 3 = 1.0986.
Ans. Let ‘x’ be the population of city.
Differential Equation
Mahesh Tutorials Science
32
∴
dx
∝x
dt
∴
dx
dt
= kx
∴
dx
dt
∴
dx
x
= kdt
= kx
Integrating both sides, we get
∫
∴
Integrating both sides, we get
dx
= k dt
x
∫
∫
log x = kt + c
∴
dx
x
∫
= k dt
log x = kt + c
When t = 0, x = X0
Initially i.e.
∴
log X0 = k (0) + c
When t = 0, x = 1000
∴
c = log X0
∴
log 1000 = k (0) + c
∴
log x = kt + log X0
∴
c = log 1000
When t = 60, x = 2X0
∴
log x = kt + log 1000
∴
log(2X0) = 60k + log X0
∴
 2X 0 
= log 
 = 60k
 X0 
∴
k=
When t = 1, x = 2000
∴
log 2000 = k (1) + log1000
∴
log 2000 – log1000 = k
 2000 
K = log 

 1000 
1
log 2
60
When x = 3X0,
∴
log(3X0) = kt + log X0
∴
log(3X0) – log X0 =
∴
60.
∴
t =
When t = 2
 5
= log  2 2  + log 1000
 
(
60. (1.0986 )
)
= log 4 2 + log 1000
log 2
= log (4000 2 )
= 95.364
= log (4000 × 1.414)
= 95.4 years
log x = log(5656)
The proportion of the country will be triple
in 95.4 years
Q-2) Th e rate of grow th of bacteria is
proportional to the number present. If
initally, there were 1000 bacteria and the
number doubles in 1 hour. find the number
of bacteria after 2
1
hours.
2
2 = 1.414]
Ans. Let ‘x’ be the numbe of bacteria present at
time ‘t’
∴
1
2
5
log x =   log 2 + log 1000
2
t
log 2
60
log 3
=t
log 2
[Take
= log 2
dx
∝x
dt
Differential Equation
= 5656
∴
x
∴
Numbers of bacteria after 2
1
hrs is 5656.
2
Q-3) The rate of growth of the population of a
city at any time t is proportional to the
size of the population. For a certain city
it is fou nd that th e cons tant of
prop ortion ality is 0 .0 4. F in d th e
population of the city after 25 years if the
intial population is 10,000.
[Take e = 2.7182]
Ans. Let ‘x’ be the population of the city at time ‘t’
Mahesh Tutorials Science
∴
dx
∝x
dt
∴
dx
dt
33
When t = 0, let m = M
∴
log M = 0 + c
∴
c = log M
= kx
When t = 10000
∴
but k is given to be 0.04.
log M = –10–4 (10000) + log M
= –1 + log M
∴
dx
dt
= 0.04x
∴
dx
x
= 0.04dt
= – log e + log M
M
= log  
e 
Integrating both sides, we get
dx
∫x
∴
∴
∫
= 0.04 dt
log x = 0.04t + c
.... (i)
∴
log (10000) = 0 + c
∴
c = log 10000
∴
log x = 0.04 t + log (1000)
∴
M M
<
e
2
∴
m<
log x = 25(0.04) + log (10000)
Q-5) The rate of growth of population is
= 1 + log (10000)
proportional to the number present. If the
population doubled in the last 25 years
= log e + log (10000)
and the present population is 1 lac, when
will the city have population 4,00,000?
= log (10000 × e)
x = 10000(2.7182)
Ans. Let ‘x’ be the population of the city at time ‘t’
= 27,182
∴
years.
Population of city after 25 years is 27,182.
∴
dx
∝x
dt
mass that instant. If α = 10–4, show that
the mass of the substance after 10,000
∴
dx
= kx
dt
years will be lass than half its value today.
∴
dx
= kdt
x
Q-4) The rate of decay of the mass of a radio
active substance at any time is α times its
Ans. Let ‘m’ be the mass of the radio active
substance at time ‘t’ years.
∴
dm
dt
∴
dm
dt
∫
dm
m
log m
Integrating both sides, we get
dx
∫x
= 10–4m
∴
–4
= 10 dt
= 10
–4
log x = kt + c
∴
log 50000 = k (0) + c
∴
c = log 50000
∫ dt
= 10–4 + c
Intially i.e,
∫
= k dt
When t = 0, x = 50000
Integrating both sides, we get
∴
M
2
Thus, the mass of the radio active substance
is less than half its value today.
When t = 25,
∴
M
e
Since e > 2
When t = 0, x = 10000
∴
M =
When t = 25, x = 10000
∴
log 10000 = 25k + log 50000
∴
log 2 = 25k
Differential Equation
Mahesh Tutorials Science
34
∴
k=
1
log 2
25
∴
When x = 400000, then
 1

 25 log ( 2)  t + log 50000


∴
log 400000=
∴
 400000 
log 
 =
 50000 
∴
t
log 2
log 8 =
25
∴
3log 2 =
∴
t
=2
25
∴
t = 50 years.
k=
 600 
1
log 

2
 X0 
When t = 8, x = 75000
∴
log 75000 = 8k + log X0
=
t
log 2
25
t
log 2
25
 600 4

= log 
 X0 
 X 0 

∴
75000
=
360000 × 360000
X03
∴
X03
=
360000 × 360000 4320000
–
750000
25
= 1728000
Thus, the population will be 4,00,000 after
50 years from present date.
X0
∴
Q-6) The bacteria culture grows at a rate
proportional to its size. After 2 hours,
= 120
Initial population is 120
When x = 200000
there are 600 bacteria and after 8 hours
the count is 75,000. Find the initial
log 200000
population and when will the population
reach 2,00,000?
∴
∴
t.
log (5) + log (120)
2
the intial population of bacteria
∴
 200000 
log 

 120 
dx
∝x
dt
∴
log 1666.66
∴
 log1666.66 
2
 = t
log 5


∴
t
dx
dt
= kx
dx
x
= kdt
Integrating both sides, we get
∫
∴
1
 600 
= t . log 
 + log120
2
 120 
=
Ans. Let ‘x’ be the number of bacteria and X0 be
∴
 600 
8
log 
 + log X 0
2
 X0 
dx
x
= 2×
=
t
log 5
2
=
t
log 5
2
3.2217
0.6990
= 2 × 4.61
∫
= k dt
log x = kt + c
When t = 0, x = X0
∴
log X0 = k (0) + c
∴
c = log X0
= 9.22 hours
∴
Q-7) A body is heated to 1100C and placed in
air at 100C. After 1 hour its temperature
is 60 0C. How much additional time is
required for it to cool to 300C?
When t = 2, x = 600
∴
log 600 = 2k + log X0
∴
log 600 = log X0 = 2k
Differential Equation
Population will reach 2,00,000 in 9.22 hours.
Ans. Let ‘x’ be the temperature of the body at time t.
∴
dx
∝ (x – 10)
dt
Mahesh Tutorials Science
35
BASIC ASSIGNMENTS
∴
dx
dt
= k.(x – 10)
∴
dx
x – 10
= k dt
Integrating both sides, we get
BA – 1
Q-1) Obtain th e d ifferen tial equation by
eliminating the arbitrary constants from
the following equations.
∫
∴
dx
x – 10
∫
i) y = c1e3x + c2e2x
= k dt
ii) y2 = (x + c)3
log|x – 10| = kt + c
When t = 0, x = 110
∴
log|110 – 10| = k(0) + c
∴
c = log (100)
Ans. i) y = c1e3x + c2e2x
Dividing throughout by e2x, we get
ye–2x = e1ex + c2
Differentiating w.r.t.x, we get
When t = 1, x = 60
∴
log|60 – 10| = k(1) + log 100
∴
 50 
k = log 

 100 
–2ye–2x + e–2x
∴
1
= log  
2
 dy

– 2y  = c
e–3x 
1
 dx

1
log|30 – 10| = t. log   + log (100)
2
∴
1
log 20 – log 100 = t. log  
2
∴
1
log   =
5
∴
t
1
t. log  
2
 – log 5 

= 
 – log 2 
=
 dy

– 2y  = c ex
e–2x 
1
 dx

Dividing throughout by ex, we get
When x = 30
∴
dy
= c1ex
dx
Again, differentiating w.r.t.x, we get
 d 2y 2dy 
 dy

– 2y  = 0
e –3x  2 –
 –3e–3x 
dx 
 dx

 dx
∴
 d 2y

dy
+ 6y  = 0
e –3x  2 – 5
dx
 dx

∴
d 2y
dy
–5
+ 6y = 0
2
dx
dx
log 5
log 2
is the required D.E
ii) y2 = (x + c)3
= 2.32
...(i)
Differentiating w.r.t.x, we get
This much time is required from starting
time.
2y
More required time after 1 hour
dy
= 3(x + c)2
dx
...(ii)
Dividing (i) by (ii), we get
 log 5

– 1
= 
 log 2

y2
dy
2y
dx
i.e. 1.32 hours.
∴
3
=
y
=
 dy 
2

 dx 
(x + c )
2
3 (x + c )
x +c
3
Differential Equation
Mahesh Tutorials Science
36
Integrating both sides, we get
3y
 dy 
2

 dx 
∴ x+c =
∫
∫
2 e x dx = 3 e –2y dy
3y
∴ c = –x +
 dy 
2

 dx 
Substituting value of c in (i), we get
3










3y 
3y  
y2 =  x +  – x +
= 

  dy  

 dy   
2

 
 2  dx  


 dx   
 


3
3e –2y
+c
–2
∴
2ex
∴
–4ex = 3e–2y + c
∴
4ex + 3e–2y = c, is the required general solution
=
Q-4) For the following differential equations
find the particular solution satisfying the
given condition.
(x – y2x) dx – (y + x2y) dy = 0;
when x = 2, y = 0.
27y
=
3
 dy 
8

 dx 
Ans. (x – y2x) dx – (y + x2y) dy = 0
3
3
 dy 
∴ 8
 = 27y
 dx 
∴
(x – y2x) dx = y(1 + x2) dy
∴
xdx
1+ x2
ydy
1 – y2
=
Integrating both sides, we get
3
 dy 
∴ 8
 – 27y = 0 is the required D.E
 dx 
Q-2) Form the differential equation of all
parabolas having the vertex at origin and
axis along the positive Y-axis.
2x
∫ 1+ x
2
dx
2x
dx
1+ x2
∴
∫
∴
log |1 + x2|
=
2y
∫1– y
= –
dy
2
–2y
∫1– y
2
dx
= –log |1 – y2| + log c
2
Ans. Let the equation of parabola be x = 4ay
∴
= log
x2
= 4a
y
Differentiating w.r.t.x, we get
d
dy
y.
x 2 – x 2.
dx
dx
=0
y2
∴
(1 + x2)
∴
(1 + x2)(1 – y2) = c
=
( )
c
1 – y2
c
(1 – y )
2
When x = 2, y = 0
∴
2xy – x2
∴
2y – x
∴
x
dy
=0
dx
dy
=0
dx
dy
– 2y = 0 is the required D.E
dx
(1 + 4)(1 – 0) = c
∴
c=5
∴
Particular solution is (1 + x2)(1 – y2) = 5
Q-5) S olve :
x=
Q-3) Solve the following differential equations.
dy x + y +1
1
=
wh en y =
at
3
dx x + y – 1
2
3
2ex+2y dx – 3dy = 0
Ans. 2ex+2y dx – 3dy = 0
∴
2ex e2y dx = 3dy
∴
2ex dx = 3.e–2y dy
Differential Equation
Ans.
dy
dx
=
x + y +1
x +y –1
Put x + y = u ∴
y=u–x
Mahesh Tutorials Science
∴
dy
dx
=
du
–1
dx
∴
du
–1
dx
=
u +1
u –1
37
Put
Differentiating w.r.t. x, we get
x
∴
du
dx
=
u +1
u +1 + u – 1
+1 =
u –1
u –1
∴
du
dx
=
2u
u –1
∴
u –1
du = 2dx
u
∴
1–
∴
x
∴
∴
1
du = 2 dx
u
dy
d
– y.
(x )
dx
dx
=
x2
dy
– y (1)
dx
x2
sin t.
=
dt
dx
dt
= ex
dx
∫ sin t dt = ∫ e
x
dx
∫1 – u du = 2∫ dx
∴
–cot t = ex + c
∴
u – log u = 2x + c
∴
y 
– cos   = e x + c
x 
∴
x + y – log(x + y) = 2x + c
∴
–log (x + y) = x – y + c
∴
y 
e x + cos   + c = 0
x
Put x =
1
2
and y =
3
3
1
+c
3
∴
–log(1) =
∴
1
c= –
3
∴
1
–log (x + y) = x – y –
3
∴
1
log (x + y) = y – x + ; is the required
3
dt
dx
Integrating both sides, we get
Integrating both sides, we get
1
y
. =1
x
This is the required particular solution
Q-2) Solve the following differential equations.
x
Ans. x
dy
y
– y + x sin 
dx
x

=0

dy
y 
– y + x sin   = 0
dx
x 
Dividing throughout by x, we get
dy y
y 
– + sin   = 0
dx x
x 
particular solution
BA – 2
 dy

y
Q-1) Solve :  x
– y  sin   = x2ex (Hint : Put
 dx

x
Put
y
=t
x
∴
y = tx
∴
dy
dt
=t +x
dx
dx
∴
t +x
∴
x
∴
dt
=
sin t
y = ux)
 dy

y 
– y  sin   = x 2e x
Ans.  x
dx


x
∴
 dy

 x dx – y 
y 
x

 sin   = e
2
x
x




dt
– t + sin t = 0
dx
dt
= –sin t
dx
–dx
x
Integrating both sides, we get
Differential Equation
Mahesh Tutorials Science
38
dx
∫ cosect dt = – ∫ x
∴
1
log (2)
3
k=
∴
log | cosec t – cot t| = –log x + log c
When t = 15
∴
y 
y 
c 
log cosec   – cot   = log  
x
x
 
 
x 
log x =
∴
y 
y  c
cosec   – cot   =
x
x x
= 5. log 2 + log X0 = log (25) + log X0
= log (32. X0)
∴
∴

y
x cosec 
x

∴

 y 
cos   

1
 x  =c
x
–
y

 
y 
 sin  x  sin  x  
 
 

∴

 y 
 – cot  x   = c

 
x
= 32 X0
Thus, the quantity of yeast becomes 32 time
the initial quantity after 15 hours.
Q-4) Find the differential equation of all circles
whose radius is 5 and centre is any point
(h, k).
Ans. Equation of circle is
(x – h)2 + (y – k)2 = 25

 y 
y 
x 1 – cos    = c sin  
 x 
x

...(i)
Differentiating w.r.t. x, we get
2(x – h) + 2(y – k) y′ = 0
This is the required general solution
∴
Q-3) In a culture of yeast the active ferment
(y – k).y′ = –(x – h)
Squaring,
doubles itself in 3 hours. Assuming that
the quan tity in creases at a rate
(y – k)2 (y′)2 = (x – h)2
(y – k)2 (y′2 = 25 – (y – k)2
...[from (i)]
proportional to itself, determine the
number of times it multiplies itself in 15
∴
(y – k)2 (y′)2 + (y – k)2 = 25
hours.
∴
( y – k)2 ((y′)2 + 1) = 25
∴
dx
∝ x
dt
∴
dx
dt
∴
dx
x
(y – k )
∴
= kx
2
d
dx
( (y ′) +1) + ((y ′) +1) dxd (y – k )
2
2
∴
(y – k)2 (y′′)2 = ((y′)2 + 1)2
∫
When t = 0, x = X0
∴
log X0 = k (0) + c
∴
c = log X0
When t = 3, x = 2X0
∴
log (2X0) = k (3) + log X0
∴
log (2X0) – log X0 = 3 k
Differential Equation
(y – k) y′′ = – ((y′)2 + 1)
Squaring,
dx
= k dx
x
log x = kt + c
=0
(y – k)2 (2(y′)(y′′)) = + ((y′)2 + 1)2(y – k).y′ = 0
(y – k) y′′ + (y′)2 + 1 = 0
= kdt
2
Dividing throughout by 2(y – k). y, we get
Integrating both sides, we get
∫
...(ii)
Differentiating w.r.t. x, we get
Ans. Let ‘x’ be the quantity of yeast present at time t.
∴
15
log 2 + log X0
3
2
25 ( y ′′ )
∴
2
(y ′ )
+1
= ((y′)2 + 1)2
∴
25(y′′)2 = ((y′)2 + 1)3
∴
  dy 2

 d 2y 
25.  2  =  
+1

  dx 

 dx 


2
3
Mahesh Tutorials Science
39
Q-5) The population grows at the rate of 8%
per year. Find the time taken for the
population to become double.
Given : log 2 = 0.6912
Ans. Let ‘x’ be the population at any time ‘t’.
8
x = 0.08x
100
∴
dx
dt
=
∴
dx
x
= 0.08 dt
Integrating both sides, we get
dx
∫x
∴
∫
= 0.08 dt
log x = 0.08t + c
Let x = X0 when t = 0
∴
log X0 = 0.08 (0) + c
∴
c = log X0
When x = 2X0,
∴
log (2X0) = t.(0.08) + log X0
∴
log (2X0) – log X0 = 0.08 (t)
∴
 2X 0 
0.08t = log 

 X0 
∴
t=
log 2
0.6912
=
≈ 8.66 years
0.08
0.08
Thus, the population doubles in 8.66 years
approximately.
Differential Equation