Question

solve the differential equation.
left( \frac { 1 } { 5 } sec x
ight) \frac { d y } { d x } = 8 e ^ { y } sin x
the solution is given by the equation
(type an equation. type an exact answer.)

Explanation:

Step1: Separate variables

We start with the differential equation \((\frac{1}{5}\sec x)\frac{dy}{dx}=8e^{y + \sin x}\).
Since \(e^{y+\sin x}=e^{y}\cdot e^{\sin x}\), the equation can be rewritten as \(\frac{1}{5}\sec x\frac{dy}{dx}=8e^{y}e^{\sin x}\).
Multiply both sides by \(5\cos x\) (because \(\sec x=\frac{1}{\cos x}\)) to get \(\frac{dy}{dx}=40e^{y}\cos x e^{\sin x}\).
Separate the variables: \(\frac{dy}{e^{y}} = 40\cos x e^{\sin x}dx\), or \(e^{-y}dy=40\cos x e^{\sin x}dx\).

Step2: Integrate both sides

Integrate \(\int e^{-y}dy=\int40\cos x e^{\sin x}dx\).
For \(\int e^{-y}dy\), using the formula \(\int e^{ax}dx=\frac{1}{a}e^{ax}+C\) (\(a = - 1\)), we have \(-e^{-y}+C_1\).
For \(\int40\cos x e^{\sin x}dx\), let \(u=\sin x\), then \(du=\cos xdx\).
So \(\int40\cos x e^{\sin x}dx = 40\int e^{u}du\).
Using the formula \(\int e^{u}du=e^{u}+C\), we get \(40e^{\sin x}+C_2\).

Step3: Solve for \(y\)

We have \(-e^{-y}=40e^{\sin x}+C\) (where \(C = C_2 - C_1\)).
Multiply both sides by \(- 1\): \(e^{-y}=-40e^{\sin x}-C\).
Take the natural logarithm of both sides: \(-y=\ln(-40e^{\sin x}-C)\).
So \(y=-\ln(-40e^{\sin x}-C)\)

Answer:

\(y =-\ln(-40e^{\sin x}+C)\)