Lecture Notes in Differential Equations - Bruce E. Shapiro

32 LESSON 4. LINEAR EQUATIONS
The equation is already given in standard form, with p(t) = −2. Hence an
integrating factor is
(∫ t
)
µ(t) = exp (−2)dt = e −2t (4.59)
0
Multiply the original ODE by the integrating factor µ(t) gives
e −2t (y ′ − 2y) = (e −2t )(e 7t ) (4.60)
Simplifying and recognizing the left hand side as the derivative of µy,
d (
ye
−2t ) = e 5t (4.61)
dt
Multiply by dt and integrate:
∫ d (
ye
−2t ) ∫
dt =
dt
e 5t dt (4.62)
From the initial condition
Hence C = 4/5 and the solution is
Example 4.5. Find the general solutions of
ye −2t = 1 5 e5t + C (4.63)
y = 1 5 e7t + Ce 2t (4.64)
1 = 1 5 e0 + Ce 0 = 1 5 + C (4.65)
y = 1 5 e7t + 4 5 e2t (4.66)
ty ′ + 4y = 4t 2 (4.67)
To solve this we first need to put it into standard linear form by dividing
by t:
y ′ + 4y = 4t2 = 4t (4.68)
t t
Since p(t) = 4/t an integrating factor is
(∫ ) 4
µ = exp
t dt = exp (4 ln t) = t 4 (4.69)