SECTION 5.5 Differential Equations 313
y
x+ 1
=
∫ xdx
(x+ 1)^2
=
∫ (x+ 1−1)dx
(x+ 1)^2
=
∫ Ñ 1
x+ 1
−
1
(x+ 1)^2
é
dx
= ln(x+ 1) +
1
x+ 1
+C
It follows, therefore, that
y = (x+ 1) ln(x+ 1) +c(x+ 1),
wherecis an arbitrary constant.
Exercises
- Solve the following first-order ODE.
(a) xy′+ 2y= 2x^2 , y(1) = 0
(b) 2x^2 y′+ 4xy=e−x, y(2) = 1.
(c) xy′+ (x−2)y= 3x^3 e−x, y(1) = 0
(d) y′lnx+
y
x
=x, y(1) = 0
(e) y′+ (cotx)y= 3 sinxcosx, y(0) = 1
(f) x(x+ 1)y′−y= 2x^2 (x+ 1), y(2) = 0
- The first-orderBernoulliODE are of the form
y′+p(x)y=q(x)yn,
wherenis any number other than 1. Show that the substitution
u=y^1 −n brings the above Bernoulli equation into the first-order
linear ODE
1
1 −n
u′+p(x)u=q(x).