1. Verify by substitution that
00
0
y1 = ex cos x, y2 = ex sin x
are solutions to
y − 2y + 2y = 0.
0
y1 = ex cos x. Then y1 = ex cos x − ex sin x and
00
0
y1 = −2ex sin x. So y1 − 2y1 + 2y1 = −2ex sin x − 2ex cos x +
2ex sin x + 2ex cos x = 0. Now lets suppose y2 = ex sin x. Then
0
00
00
0
y2 = ex sin x + ex cos x and y2 = 2ex cos x. So y2 − 2y2 + 2y2 =
2ex cos x − 2ex sin x − 2ex cos x + 2ex sin x = 0. Therefore, this
00
0
shows that y1 , y2 are solutions to y − 2y + 2y = 0
Solution:
Suppose
00
2. Verify by substitution that
Solution:
Suppose
y=
1
1+x2 .
y =
1
1+x2
0
is a solution to
y + 2xy 2 = 0.
0
Then
2x
y = − (1+x)
2 = −2x
1
1+x2
2
=
0
3. Find a
−2xy 2 =⇒ y + 2xy 2 = 0
√
dy
2
solution to
dx = x x + 4 such
dy
dx
Solution:
y (−4) = 0.
´ dy
´ √
y (x) = dx
dx = x x2 + 4dx. If we let u
√
´
´
x2 + 4, then du = 2xdx. So x x2 + 4dx = 12 u1/2 du
3/2
1 3/2
+ C = 13 x2 + 4
+ C . This implies that y (x)
3u
3/2
3/2
1
2
+ C . Since we have 0 = y (−4) = 13 (20)
3 x +4
√
√
3/2
C =⇒ C = − 20
20. So y (x) = 13 x2 + 4
20
− 20
3
3
Solution:
4. Solve
that
Observe that
=
ex
y2
such that
dy
dx
x
=
ex
y2
=
=
=
+
y (0) = 2.
1 3
3y
x
=⇒ y 2 dy = ex dx =⇒
= ex + C =⇒ y 3 =
1/3
3e + 3C = 3ex + C =⇒ y (x) = (3e + C)
.
Since we have
(3ex + 5)
5. Solve
dy
dx
Solution:
1/3
1/3
2 = y (0) = (3 + C)
=⇒ C = 5.
So
y (x) =
.
= (1 − y) cos x
such that
dy
dx
y (π) = 2.
dy
= (1 − y) cos x =⇒ 1−y
= cos xdx =⇒ ln |1 − y| = sin x +
sin x
C =⇒ 1 − y = Ce
=⇒ y (x) = 1 + Cesin x .
1