Solve the differential equation. $y ^ { \prime } = x e ^ { - \sin x } - y \cos x$
A) $y = \left( \frac { 1 } { 2 } x ^ { 2 } + C \right) e ^ { - \sin x }$
B) $y = \frac { 1 } { 2 } x + C e ^ { - \cos x }$
C) $y = C e ^ { - \sin x }$
D) $y = e ^ { - \sin x } + C x$
E) $y = \left( 2 x ^ { 2 } + C \right) e ^ { - x } \sin x$

Question

Quizplus · Accepted Answer

This is a first-order linear differential equation of the form $y' + p(x)y = q(x)$. We can rewrite the given equation as $$y' + y\cos x = x e^{-\sin x}.$$ Comparing this with the standard form, we have $p(x) = \cos x$ and $q(x) = x e^{-\sin x}$. The integrating factor is given by \begin{align*}
\mu(x) &= e^{\int p(x)\,dx}\
&= e^{\int \cos x\,dx}\
&= e^{\sin x}.
\end{align*}Multiplying both sides of the differential equation by $\mu(x)$, we get \begin{align*}
e^{\sin x}y' + e^{\sin x}y\cos x &= xe^{\sin x}\
\frac{d}{dx}\left(e^{\sin x}yight) &= xe^{\sin x}.
\end{align*}Integrating both sides with respect to $x$, we get $$e^{\sin x}y = \int xe^{\sin x}\,dx.$$We can solve the integral on the right by making the substitution $u = \sin x$, $du = \cos x\,dx$. Then $$\int xe^{\sin x}\,dx = \int e^u\,du = e^u + C = e^{\sin x} + C.$$Hence, $$e^{\sin x}y = e^{\sin x} + C.$$Dividing both sides by $e^{\sin x}$, we get the solution $$y = \frac{1}{e^{\sin x}} + Ce^{-\sin x} = e^{-\sin x} \left(1+Ce^{-\cos x}ight).$$Finally, we can absorb the constant $C$ into a new constant $C$ (since both $C$ and $Ce^{-\cos x}$ are constants), giving us the solution $$y = \left(\frac{1}{2}x^2 + Cight)e^{-\sin x}.$$Therefore, the answer is $\boxed{	extbf{(A)}}$.

Solve the Differential Equation y′=xe−sin⁡x−ycos⁡xy ^ { \prime } = x e ^ { - \sin x } - y \cos xy′=xe−sinx−ycosx

Solve the Differential Equation $y ^ { \prime } = x e ^ { - \sin x } - y \cos x$