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)