Lecture Notes in Differential Equations - Bruce E. Shapiro

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Example 4.1. Solve the differential equation
y ′ + 2ty = t (4.29)
This has the form y ′ + p(t)y = q(t), where p(t) = 2t and q(t) = t. An
integrating factor is
(∫ ) (∫ )
µ(t) = exp p(t)dt = exp 2tdt = e t2 (4.30)
Multiplying equation (4.29) through by the integrating factor µ(t) gives
e t2 (y ′ + 2ty) = te t2 (4.31)
Recall that the left hand side will always end up as the derivative of yµ
after multiplying through by µ; we can also verify this with the product
rule:
(
d
ye t2) = y ′ e t2 + 2te t2 y = e t2 (y ′ + 2ty) (4.32)
dt
Comparing the last two equations tells us that
d
dt
(
ye t2) = te t2 (4.33)
Multiply through both sides by dt and integrate:
∫ ( d
ye t2) ∫
dt = te t2 dt (4.34)
dt
The left hand side is an exact derivative, hence
∫
ye t2 = te t2 dt + C (4.35)
The right hand side can be solved with the substitution u = t 2 :
∫
te t2 dt = 1 ∫
e u du = 1 2 2 eu + C = 1 2 et2 + C (4.36)
Combining the last two equations,
Hence we can solve for y,
ye t2 = 1 2 et2 + C (4.37)
y = 1 2 + Ce−t2 . (4.38)