Question

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

Answer 1

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 $C$ :

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

Then the particular solution to the initial value problem is

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

(A)