Problem 5.13
Problem. Solve the differential equation xy ′′ − (x + 2)y ′ + 2y = 0.
Solution.
Use the substitution y = ex v. Then
y ′ = ex (v ′ + v),
y ′′ = ex (v ′′ + 2v ′ + v).
Substitution gives
0 = xex (v ′′ + 2v ′ + v) − (x + 2)ex (v ′ + v) + 2ex v
= ex xv ′′ + (x − 2)v ′ .
Thus
xv ′′ + (x − 2)v ′ = 0.
Let u = v ′ . On an interval where x ̸= 0,
xu′ + (x − 2)u = 0,
u′
2
= − 1,
u
x
ln |u| = 2 ln |x| − x + C,
u = C2 x2 e−x .
Therefore
Z
v = C 1 + C2
x2 e−x dx
= C1 − C2 e−x (x2 + 2x + 2).
Absorbing the minus sign into C2 ,
y = ex v = C1 ex + C2 (x2 + 2x + 2).
The standard largest intervals are (−inf ty, 0) and (0, inf ty); these solution functions also extend
through x = 0.
Final answer.
y = C1 ex + C2 (x2 + 2x + 2)
Confidence: high
1