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

Question

Quizplus · Accepted Answer

This differential equation can be solved by recognizing it as a first-order linear differential equation in the form $y' + P(x)y = Q(x)$, where $P(x) = \cos x$ and $Q(x) = xe^{-\sin x}$. The integrating factor is $e^{\int P(x)dx} = e^{\int \cos x dx} = e^{\sin x}$. Multiplying both sides by this integrating factor and integrating with respect to $x$ gives the solution $y = \left( \frac { 1 } { 2 } x ^ { 2 } + C \right) e ^ { - \sin x }$.

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

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