Worksheet 6 - Solutions
1. Find a solution to the initial value problem
dy
= ey x3
dx
y(0) = 0
Solution: In what follows, the value of the constant of integration
may change from line to line.
dy
= e y x3
dx
e−y dy = x3 dx
ˆ
ˆ
e−y dy = x3 dx
1
−e−y = x4 + C
4
1
e−y = − x4 + C
4
1
y = − ln(C − x4 )
4
Substituting the initial condition 0 = y(0) = − ln(C), we find that
C = 1 and y(x) = − ln(1 − 41 x4 )
2. Find a solution to the initial value problem
dy
= (1 + y 2 )ex
dx
y(0) = 0
Solution:
dy
= (1 + y 2 )ex
dx
dy
= ex dx
1 + y2
ˆ
ˆ
dy
= ex dx
1 + y2
arctan(y) = ex + C
y = tan(ex + C)
1
Substituting in the initial condition, we find that 0 = Y (0) = tan(1 +
C). A possible choice of C is C = −1. Our final answer is then
y(x) = tan(ex − 1).
3. Find a solution to the initial value problem
p
dy
= y y 2 − 1 cos(x)
dx
y(0) = 1
Solution: First, we can observe that one solution to this problem is
given by y(x) = 1.
We can find another solution by separating variables.
p
dy
= y y 2 − 1 cos(x)
dx
dy
p
= cos(x)dx
y y2 − 1
ˆ
ˆ
dy
p
= cos(x)dx
y y2 − 1
arcsec(y) = sin(x) + C
y = sec(sin(x) + C)
Substituting in the initial condition y(0) = 1 we find that
1 = y(0) = sec(C)
So we may take, for example, C = 0. Our final solution is then either
of y(x) = 1 or y(x) = sec(sin(x)).
4. Find the general solution to the differential equation
dy
= x2 + y 2 x2
dx
2
Solution:
dy
dx
dy
dx
dy
1 + y2
ˆ
dy
1 + y2
= x2 + y 2 x2
= x2 (1 + y 2 )
= x2 dx
ˆ
= x2 dx
x3
+C
3 x3
y(x) = tan
+C
3
arctan(y) =
.
5. Find the general solution to the differential equation
1
dy
= √
y
dx
e 1 − x2
Solution:
dy
1
= √
y
dx
e 1 − x2
dx
ey dy = √
2
ˆ
ˆ 1−x
dx
√
ey dy =
1 − x2
y
e = arcsin(x) + C
y = ln (arcsin(x) + C)
6. Find the general solution to the differential equation
dy
1
= y
dx
e (1 + x2 )
3
Solution:
1
dy
= y
dx
e (1 + x2 )
dx
ey dy =
(1 + x2 )
ˆ
ˆ
dx
ey dy =
(1 + x2 )
ey = arctan(x) + C
y = ln (arctan(x) + C)
7. Find a solution to the initial value problem
p
dy
= 1 − y 2 sec2 (x)
dx
y(0) = 0
Solution:
ˆ
1
p
1 − y2
p
dy
= 1 − y 2 sec2 (x)
dx ˆ
dy = sec2 (x)dx
arcsin(y) = tan(x) + C
y(x) = sin(tan(x) + C)
Using the initial condition, we find that
0 = sin(0 + C)
so C = 0 gives a solution. Our final answer is then sin(tan(x)).
8. Find the general solution to the differential equation (for x 6= 0)
x
dy
= −y + x
dx
4
Solution: We rewrite the equation as
dy
x
+y =x
dx
and observe that this equation is already in the form
d(xy)
=x
dx
which is separable. We solve
d(xy)
=x
ˆ dx
ˆ
d(xy) = xdx
1
xy = x2 + C
2
C
1
y(x) = x +
2
x
9. Find the general solution to the differential equation
1 dy
2
= y + ex
2x dx
Solution: We begin by writing the problem in standard form as
dy
2
− 2xy = 2xex
dx
´
2
The integrating factor for this problem is m(x) = e −2xdx = e−x .
2
If we multiply through by e−x , then the equation becomes separable
and we can find the general solution directly directly.
2 dy
2
e−x
− 2xe−x y = 2x
dx
2
d(e−x y)
= 2x
dx
ˆ
ˆ
2
d(e−x y) =
2xdx
2
e−x y = x2 + C
2
y(x) = x2 ex + Cex
5
2
10. Find a solution to the initial value problem
x
dy
cos(x)
+ 2y =
dx
x
y(π) = 1
Solution: We being by writing the differential equation in standard
form as
dy
2
cos(x)
+ y=
dx x
x2
´
2
The integrating factor for this problem is m(x) = e x dx = e2 ln(x) =
x2 . Multiplying through by x2 converts this problem to
x2
dy
+ 2xy = cos(x)
dx
d(x2 y)
= cos(x)
ˆ dx
ˆ
2
d(x y) = cos(x)dx
x2 y = sin(x) + C
sin(x)
C
y(x) =
+ 2
x2
x
We substitute in the initial condition y(π) = 1 to find that
y(π) =
sin(π) C
+ 2
2
| π
{z } π
0
so C = π 2 and y(x) =
sin(x)
x2
+
π2
.
x2
6
11. Find a solution to the initial value problem
dy
= 1 − sin(x)y
dx
y(0) = 1
cos(x)
Solution: We begin by writing the equation in standard form
dy
+ tan(x)y = sec(x)
dx
´
The integrating factor for this problem is m(x) = e tan(x)dx = e− ln(cos(x)) =
sec(x). Multiplying through by m(x) makes this equation separable.
dy
+ tan(x)y = sec(x)
dx
sec(x)
dy
+ sec(x) tan(x)y = sec2 (x)
dx
d(sec(x)y)
= sec2 (x)
dx
ˆ
ˆ
d(sec(x)y) = sec2 (x)dx
sec(x)y = tan(x) + C
y = sin(x) + C cos(x)
Substituting in y(0) = 1 we find that C = 1 and y(x) = sin(x)+cos(x).
12. Find a solution to the initial value problem
x
dy
sin(x)
+ 2y = −
dx
x
π
y( ) = 1
2
Solution: We being by writing the differential equation in standard
form as
dy
2
sin(x)
+ y=− 2
dx x
x
7
´
2
The integrating factor for this problem is m(x) = e x dx = e2 ln(x) =
x2 . Multiplying through by x2 converts this problem to
dy
x2
+ 2xy = − sin(x)
dx
d(x2 y)
= − sin(x)
ˆ dx
ˆ
d(x2 y) = −
sin(x)dx
x2 y = cos(x) + C
cos(x)
C
y(x) =
+ 2
2
x
x
Susbtituting in the initial condition, we find that
cos( π2 )
π
C
1 = y( ) = π 2 + π 2
2
(2)
(2)
| {z }
0
so that y(x) =
cos(x)
x2
+
π2 1
4 x2 .
13. Find a solution to the initial value problem
1
dy
= (y − 1)
dx
x
y(−1) = 0
Solution:
dy
1
= (y − 1)
dx
x
1 dy
1
=
y − 1 dx
x
1
1
dy = dx
y−1
x
ˆ
ˆ
1
1
dy =
dx
y−1
x
ln |y − 1| = ln|x| + c
y − 1 = ±|x|ec
y = 1 ± |x|ec
8
We are working near −1, so |x| = −x. Plugging in y(−1) = 0,
0 = 1 ± ec (−(−1))
| {z }
| − 1|
ec is always positive, so we must have
0 = y(−1) = 1 − ec
Thus 1 = ec and we get as our final answer
y(x) = 1 − ec (−x)
y(x) = 1 + x
14. Find the general solution to the differential equation
cos(x)
where we assume that
−π
2
dy
= y + sin(x) + 1
dx
< x < π2 .
Solution: We begin by writing the differential equation in standard
form
cos(x)
dy
= y + sin(x) + 1
dx
dy
− sec(x)y = tan(x) + sec(x)
dx
´
The integrating factor is m(x) = e − sec(x)dx = e− ln | sec(x)+tan(x)| . Re1
calling that we assumed − π2 < x < π2 , this is sec(x)+tan(x)
. Multiplying
through, we find that
d
y
=1
dx sec(x) + tan(x)
ˆ
ˆ
y
d
= dx
sec(x) + tan(x)
y
=x+C
sec(x) + tan(x)
y(x) = x (sec(x) + tan(x)) + C (sec(x) + tan(x))
9
15. Find the general solution to the differential equation
dy
1
3√
+ 2
y=
1+x
dx x − 1
2
where we assume that x > 1.
Solution: The equation is already in standard form,
can solve
q so we q
´ 1
ln | x−1
|
dx
2
x+1
for the integrating factor m(x) = e x −1 = e
= x−1
x+1 for
x > 1. Multiplying through, we find that
r
r
r
x − 1 dy
1
x−1
3 x − 1√
1+x
+
y=
x + 1 dx x2 − 1 x + 1
2 x+1
r
d
x−1
3√
y
=
x−1
dx
x+1
2
!
r
ˆ
ˆ
3 √
x−1
d y
=
x − 1dx
x+1
2
r
3
x−1
= (x − 1) 2 + C
y
x+1
r
√
x+1
y(x) = (x − 1) x + 1 + C
x−1
16. Find a solution to the initial value problem
x2
dy
− 2xy = x4 cos(x)
dx
y(π) = 1
Solution: We begin by putting the differential equation into standard
form
x2
dy
− 2xy = x4 cos(x)
dx
dy
2
− y = x2 cos(x)
dx x
10
The integrating factor is then m(x) = e−2
multiply through to find that
´
1
dx
x
= e− ln |x| =
1
.
x2
We
dy
2
− y = x2 cos(x)
dx x
1 dy
2
− y = cos(x)
x2 dx x3
1
d
y = cos(x)
dx x2
ˆ
ˆ 1
d
y = cos(x)dx
x2
1
y = sin(x) + C
x2
y(x) = x2 sin(x) + Cx2
Substituting in the initial condition, we find that
1 = y(π) = π 2 sin(π) + Cπ 2 = Cπ 2
so that C =
1
π2
and y(x) = x2 sin(x) +
x2
.
π2
17. Find a solution to the initial value problem
dy
= (1 + x2 )ex − y
dx
y(tan(1)) = etan(1)
(1 + x2 )arctan(x)
Solution: We begin by putting the equation into standard form
dy
1
ex
+
y
=
dx (1 + x2 )arctan(x)
arctan(x)
Notice that
ˆ
ˆ
1
1
dx =
du
2
arctan(x)(1 + x )
u
= ln |u| + C
u = arctan(x)
du =
1
dx
1 + x2
= ln |arctan(x)| + C
We are working near tan(1) > 0 so we may assume that arctan(x) > 0.
11
´
1
dx
m(x) = e arctan(x)(1+x2 )
= eln(arctan(x))
= arctan(x)
Multiplying through by the integrating factor, we find that
d
(arctan(x)y) = ex
dx
ˆ
ˆ
d (arctan(x)y) =
ex dx
arctan(x)y = ex + C
C
ex
+
y(x) =
arctan(x) arctan(x)
Substituting in the initial condition
etan(1) = y(tan(1)) =
C
etan(1)
+
arctan(tan(1)) arctan(tan(1))
= etan(1) + C
x
e
so C = 0 and y(x) = arctan
.
(x)
18. Find a particular solution to the differential equation
1 + x3 dy
= 1 − y(x)
3x2 dx
y(1) = 2
Solution: We first put the equation into standard form
dy
3x2
3x2
+
y
=
dx 1 + x3
1 + x3
12
The integrating factor for this problem is m(x) = (1 + x3 ) and the
solution is
ˆ
1
2
y(x) =
3x dx
1 + x3
1
x3 + C
=
3
1+x
The initial condition y(1) = 2 gives that
3
.
y(x) = 3+x
1+x3
13
1+C
2
= 2, so C = 3 and