MATH/MTHE 225 - WINTER 2017
Problem Set 5
5xy + 18x2 + 2y 2
1. Solve the differential equation: 5y =
.
x2
0
2. Solve the initial value problem: y 0 =
6y 2 + 5x2
,
xy
y(1) = 1.
7
7
3. Solve the differential equation: y 0 − y = − e−7x y 9 .
8
8
4. Solve the differential equation: y 0 +
3
12
y = 96x4 y 4 .
x
5. Solve the differential equation: y 0 = sin2 (x − y).
6. Solve the differential equation: y 0 = (6x − 2y + 1)2/3 + 3.
1
Solution
1.
Notice: 5y 0 =
5xy + 18x2 + 2y 2
5xy + 18x2 + 2y 2
dy
=
⇒
5
x2
dx
x2
⇒ 5x2 dy − (5xy + 18x2 + 2y 2 ) dx = 0
(?)
This differential equation“looks like” either as a homogeneous equation or an exact equation.
So let us first check if it is a homogeneous equation.
Assuming it is in homogeneous form, we have M (x, y) = −(5xy + 18x2 + 2y 2 ) and N (x, y) =
5x2 . Now notice:
M (tx, ty) = −(5(tx)(ty) + 18(tx)2 + 2(ty)2 ) = t2 [−(5xy + 18x2 + 2y 2 )] = t2 M (x, y)
N (tx, ty) = 5(tx)2 = t2 (x2 ) = t2 N (x, y)
Hence the given differential equation is a homogeneous equation.
Let y = ux. That means dy = x du + u dx. Now substituting these in (?):
5x2 (x du + u dx) − (5x(ux) + 18x2 + 2(ux)2 ) dx = 0 ← divide both sides by x2
⇒ 5(x du + u dx) − (5u + 18 + 2u2 ) dx = 0
⇒ 5x du + 5u dx − 5u dx − 18 dx − 2u2 dx = 0
⇒ 5x du − 18 dx − 2u2 dx = 0
⇒ 5x du = (18 + 2u2 ) dx
5
⇒ x du = (9 + u2 ) dx
2
du
2 dx
⇒
=
2
9+u
5 x
Z
Z
du
2
dx
⇒
=
2
9+u
5
x
1
u
2
⇒ arctan
= ln x + C
3
3
5
u 6
⇒ arctan
= ln x + C
3
5
y 6
⇒ arctan
= ln x + C
3x
5
y
6
⇒
= tan
ln x + C
3x
5
(implicit solution)
2
⇒ y = 3x tan
2.
Notice: y 0 =
6
ln x + C
5
(explicit solution)
6y 2 + 5x2
dy
6y 2 + 5x2
⇒
=
⇒ xy dy − (6y 2 + 5x2 ) dx = 0
xy
dx
xy
(?)
This differential equation“looks like” either as a homogeneous equation or an exact equation.
So let us first check if it is a homogeneous equation.
Assuming it is in homogeneous form, we have M (x, y) = −(6y 2 + 5x2 ) and N (x, y) = xy.
Now notice:
M (tx, ty) = −(6(ty)2 + 5(tx)2 ) = t2 (−6y 2 − 5x2 ) = t2 [−(6y 2 + 5x2 )] = t2 M (x, y)
N (tx, ty) = (tx)(ty) = t2 (xy) = t2 N (x, y)
Hence the given differential equation is a homogeneous equation.
Let y = ux. That means dy = x du + u dx. Now substituting these in (?):
(x)(ux)(x du + u dx) − (6(ux)2 + 5x2 ) dx = 0 ← divide both sides by x2
⇒ u(x du + u dx) − (6u2 + 5) dx = 0
⇒ ux du + u2 dx − 6u2 dx − 5 dx = 0
⇒ ux du − 5u2 dx − 5 dx = 0
⇒ ux du = 5(1 + u2 ) dx
u
dx
du = 5
2
1+u
x
Z
Z
dx
u
du = 5
(on the left integral use v−substitution with v = 1 + u2 )
⇒
2
1+u
x
1
⇒ ln (1 + u2 ) = 5 ln x + C
2
1
y2
⇒ ln 1 + 2 = 5 ln x + C
(implicit solution)
2
x
y2
⇒ ln 1 + 2 = 2(5 ln x + C)
x
10
y2
⇒ 1 + 2 = e2(5 ln x+C) = e10 ln x e2C = eln x D = x10 D;
(D = e2C )
x
⇒
⇒ x2 + y 2 = x12 D (by multiplying both sides by x2 )
⇒ y 2 = x12 D − x2
3
√
√
⇒ y = x12 D − x2 = x x10 D − 1
square root.
But we want y(1) = 1. That
means 1 =
√
value problem is: y = x 2x10 − 1.
(explicit solution) - just considering positive
√
D − 1 ⇒ D = 2. Hence the solution of the initial
3.
This differential equation clearly a Bernoulli’s differential equation with n = 9. Therefore
du
1
dy
= − u−9/8
. Now substitute all these in
set u = y 1−n = y −8 or y = u−1/8 and hence
dx
8
dx
the given differential equation:
7
7
y 0 − y = − e−7x y 9
8
8
1
du 7 −1/8
7
⇒ − u−9/8
− u
= − e−7x u−9/8 (now multiply both sides by −8u9/8 )
8
dx 8
8
du
⇒
+ 7 u = 7e−7x
(?)
dx |{z}
p(x)
This is a linear differential equation
with p(x) = 7 and hence integrating factor method is
R
7 dx
= e7x . Now multiply both sides of (?) by this I.F. and
applicable. Therefore, I.F. = e
we get:
e7x
du
+ 7e7x = 7
dx
d
⇒
(ue7x ) = 7 ⇒ d(ue7x ) = 7 dx ⇒
dx
Z
7x
d(ue ) = 7
Z
dx
⇒ ue7x = 7x + C
⇒ y −8 e7x = 7x + C
e7x
7x + C
7x 1/8
e
⇒y=
7x + C
(implicit solution)
⇒ y8 =
(explicit solution)
4.
This differential equation clearly a Bernoulli’s differential equation with n = 3/4. Theredy
du
fore set u = y 1−n = y 1/4 or y = u4 and hence
= 4u3
. Now substitute all these in the
dx
dx
given differential equation:
y0 +
3
12
y = 96x4 y 4
x
4
du 12 4
1
+ u = 96x4 u3 (now multiply both sides by u−3 )
dx
x
4
du
3
⇒
+
u = 24x4
(? ?)
dx |{z}
x
⇒ 4u3
p(x)
This is a linear differential equation
with p(x) = x3 and hence integrating factor method is
R 3
applicable. Therefore, I.F. = e x dx = e3 ln x = x3 . Now multiply both sides of (? ?) by this
I.F. and we get:
du
+ 3x2 u = 24x7
dx
Z
Z
d
3
7
3
7
3
⇒
(ux ) = 24x ⇒ d(ux ) = 24x dx ⇒
d(ux ) = 24 x7 dx ⇒ ux3 = 3x8 + C
dx
x3
⇒ y 1/4 x3 = 3x8 + C
(implicit solution)
⇒ y 1/4 = 3x5 + Cx−3
⇒ y = (3x5 + Cx−3 )4
(implicit solution)
5.
dy
= f (ax + by + c).
dx
du
dy
dy
du
Let u = x − y. That means
=1−
⇒
=1−
. Now substituting these in the
dx
dx
dx
dx
differential equation:
dy
= sin2 (x − y)
y0 =
dx
du
⇒1−
= sin2 u
dx
du
= 1 − sin2 u
⇒
dx
du
⇒
= cos2 u
dx
du
⇒
= dx
cos2 u
We see that the differential equation is of form
⇒ sec2 u du = dx
Z
Z
2
⇒ sec u du =
dx
⇒ tan u = x + C
5
⇒ tan (x − y) = x + C
(implicit solution)
⇒ x − y = arctan (x + C)
⇒ y = x − arctan (x + C)
(explicit solution)
6.
dy
= f (ax + by + c).
dx
.
dy
dy
1
du
1 du
du
=6−2
⇒
=
6−
=3−
. Now
Let u = 6x − 2y + 1. That means
dx
dx
dx
2
dx
2 dx
substituting these in the differential equation:
y 0 = (6x − 2y + 1)2/3 + 3
We see that the differential equation is of form
1 du
= u2/3 + 3
2 dx
1 du
= u2/3
⇒−
2 dx
du
⇒ 2/3 = −2 dx
u
⇒3−
⇒ u−2/3 du = −2 dx
Z
Z
−2/3
⇒ u
du = −2 dx
⇒
u1/3
1
3
= −2x + C
1
⇒ u1/3 = (−2x + C)
3
1
⇒ (6x − 2y + 1)1/3 = (−2x + C)
(implicit solution)
3
3
1
⇒ (6x − 2y + 1) = (−2x + C)
3
3
1
⇒ 2y = 6x + 1 − (−2x + C)
3
1
1
⇒ y = 3x + − (−2x + C)3
(explicit solution)
2 54
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