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gladu [14]
1 year ago
9

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

Mathematics
1 answer:
hodyreva [135]1 year ago
4 0

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)

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