Solution Techniques for Elementary Partial Differential Equations, Third Edition, Instructor’s Solutions Manual ch1-ch14
Chapter 1
Section 1.1
1. This is a variables separable equation and y = 0 is a solution. If y 6= 0, then
dy
y
=
2x
x 2 + 1
dx
⇒ ln|y| = ln(x 2 + 1) + C 0
⇒ y(x) = C(x 2 + 1),
where C 6= 0. The value C = 0 generates the singular solution y = 0.
2. Proceeding as in 1, for y 6= −1 we have
y ′ = 3x 2 (y + 1)
⇒
dy
y + 1
= 3x 2
⇒ ln|y + 1| = x 3 + C 0
⇒ y = Ce x
3
− 1,
where C 6= 0. The value C = 0 generates the singular solution y = −1.
3. This is a linear equation, solved by the integrating factor method:
dy
dx
+
2
x − 1
y =
x
x − 1
⇒ µ = exp
?Z
2
x − 1
dx
?
= e 2ln|x−1| = (x − 1) 2
⇒ y(x) =
1
(x − 1) 2
Z
(x − 1) 2
x
x − 1
dx =
1
(x − 1) 2
Z
(x 2 − x)dx
= (x − 1) −2 ( 1
3
x 3 −
1
2
x 2 + C ) .
4. Proceeding as in 3, we have
dy
dx
−
2
x
y = x 3 e x
⇒ µ = exp
?Z
−
2
x
dx
?
= e −lnx
2
=
1
x 2
⇒ y(x) = x 2
Z
1
x 2
x 3 e x dx = x 2
Z
xe x dx
= x 2
?
xe x −
Z
e x dx
?
= x 2 (xe x − e x + C).
