Solve the differential equation.
-$$e ^ { x } \frac { d y } { d x } + 3 e ^ { x } y = 2 , x 0$$
A) $y = e ^ { - x } + C e ^ { - 3 x } , x 0$
B) $y = e ^ { - 3 x } + C e ^ { - x } , x 0$
C) $y = e ^ { - x } + e ^ { - 3 x } , x 0$
D) $y = e ^ { x } + C e ^ { - 3 x } , x 0$

Question

Solve the differential equation.
-$$e ^ { x } \frac { d y } { d x } + 3 e ^ { x } y = 2 , x > 0$$
A) $y = e ^ { - x } + C e ^ { - 3 x } , x > 0$
B) $y = e ^ { - 3 x } + C e ^ { - x } , x > 0$
C) $y = e ^ { - x } + e ^ { - 3 x } , x > 0$
D) $y = e ^ { x } + C e ^ { - 3 x } , x > 0$

Quizplus · Accepted Answer

This is a first-order linear differential equation. To solve it, we can use the integrating factor method. The integrating factor is $e^{\int P(x) dx}$, where $P(x)$ is the coefficient of $y$ in the differential equation. Here, $P(x) = 3e^x$, so the integrating factor is $e^{\int 3e^x dx} = e^{3e^x}$. However, since the equation is already in a form where the left side is the derivative of $y \cdot e^{3x}$, we can directly integrate both sides with respect to $x$. The solution involves finding the general solution, which includes a constant $C$, and the particular solution. The correct solution is $y = e^{-x} + Ce^{-3x}$, which matches option A.

Solve the Differential Equation exdydx+3exy=2,x>0e ^ { x } \frac { d y } { d x } + 3 e ^ { x } y = 2 , x > 0exdxdy​+3exy=2,x>0

Solve the Differential Equation $e ^ { x } \frac { d y } { d x } + 3 e ^ { x } y = 2 , x > 0$