$\text { Solve the first-order linear differential equation } \frac { d y } { d x } = ( y - 1 ) \sin ( 2 x ) \text {. }$
A) $y = 1 + C e ^ { \sin ( 2 x ) }$
B) $y = 1 + C e ^ -{ \frac { \cos ( 2 x ) } { 2 } }$
C) $y = 1 + C e ^ { \frac { \sin ( 2 x ) } { 2 } }$
D) $y = 1 + C e ^- { \frac { \sin ( 2 x ) } { 2 } }$
E) $y = 1 + C e ^ { \cos ( 2 x ) }$

Question

$\text { Solve the first-order linear differential equation } \frac { d y } { d x } = ( y - 1 ) \sin ( 2 x ) \text {. }$ 
A) $y = 1 + C e ^ { \sin ( 2 x ) }$
B) $y = 1 + C e ^ -{ \frac { \cos ( 2 x ) } { 2 } }$
C) $y = 1 + C e ^ { \frac { \sin ( 2 x ) } { 2 } }$
D) $y = 1 + C e ^- { \frac { \sin ( 2 x ) } { 2 } }$
E) $y = 1 + C e ^ { \cos ( 2 x ) }$

Quizplus · Accepted Answer

The given differential equation is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -\sin(2x)$ and $Q(x) = \sin(2x)$. The integrating factor, $e^{\int P(x)dx}$, is $e^{-\int \sin(2x)dx} = e^{\frac{-\cos(2x)}{2}}$. Multiplying both sides of the differential equation by the integrating factor and integrating yields the solution $y = 1 + Ce^{-\frac{\cos(2x)}{2}}$.

Solve the first-order linear differential equation dydx=(y−1)sin⁡(2x). \text { Solve the first-order linear differential equation } \frac { d y } { d x } = ( y - 1 ) \sin ( 2 x ) \text {. } Solve the first-order linear differential equation dxdy​=(y−1)sin(2x).

$\text { Solve the first-order linear differential equation } \frac { d y } { d x } = ( y - 1 ) \sin ( 2 x ) \text {. }$