Math 225 Differential Equations Notes Chapter 2

Example 2: Solve the Initial value problem
Solution:
• dy
y−1 = dx
x+3
• ∫ dy
y−1 = ∫ dx
x+3
• ln|y − 1| = ln|x + 3| + C
dy
dx = y − 1
x + 3
y(−1) = 0
• now we have an implicit solution
• Let’s solve for y explicitly
• e ln|y−1| = e ln|x+3|+C
• |y − 1| = e C |x + 3| = C 1 |x + 3| where C 1 = e C > 0
• y − 1 = ±C 1 (x + 3) implying y = 1 ± C 1 (x + 3)
• y = 1 + K(x + 3) If we let K := ±C 1 and non-zero
We can solve for K using the initial condition of y(−1) = 0
0 = 1 + K(−1 + 3) = 1 + 2K
So k = −1/2 and the solution to the IVP is
y = 1 − 1 2
−1
(x + 3) = (x + 1)
2
4