From a practice assignment: solve the following differential equation given initial conditions

Question

asked Feb 22, 2023 138k views

1 Answer

Delphifirst · Answer 1 · 2023-02-26T23:05:24+0000

If
$y' = e^y \sin(x)$ and
$y(-\pi)=0$ , separate variables in the differential equation to get

$e^(-y) \, dy = \sin(x) \, dx$

Integrate both sides:

$\displaystyle \int e^(-y) \, dy = \int \sin(x) \, dx \implies -e^(-y) = -\cos(x) + C$

Use the initial condition to solve for
:

$-e^(-0) = -\cos(-\pi) + C \implies -1 = 1 + C \implies C = -2$

Then the particular solution to the initial value problem is

$-e^(-y) = -\cos(x) - 2 \implies e^(-y) = \cos(x) + 2$

(A)