Solve the differential equation.
-$$4 y ^ { \prime } = e ^ { x / 4 } + y$$
A) $y = \frac { x e ^ { x / 4 } + C e ^ { x / 4 } } { 4 }$
B) $y = \frac { x e ^ { x / 4 } + C } { 4 }$
C) $y = \frac { - x e ^ { x / 4 } + C e ^ { x / 4 } } { 4 }$
D) $y = x e ^ { x / 4 } + C e ^ { x / 4 }$

Question

Quizplus · Accepted Answer

To solve the differential equation, we can use the method of integrating factor. First, we divide both sides by 4 to get:
$$y' - \frac{1}{4}y = \frac{1}{4}e^{x/4}$$
The integrating factor is:
$$I = e^{\int -\frac{1}{4} dx} = e^{-\frac{1}{4}x}$$
Multiplying both sides by $I$ gives:
$$e^{-\frac{1}{4}x}y' -\frac{1}{4}e^{-\frac{1}{4}x}y = \frac{1}{4}$$
The left hand side can be written as:
$$\frac{d}{dx}(e^{-\frac{1}{4}x}y) = \frac{1}{4}$$
Integrating both sides with respect to $x$ gives:
$$e^{-\frac{1}{4}x}y = \int\frac{1}{4}dx + C$$
$$e^{-\frac{1}{4}x}y = \frac{1}{4}x + C$$
Multiplying both sides by $e^{\frac{1}{4}x}$ gives:
$$y = \frac{1}{4}xe^{\frac{1}{4}x} + Ce^{\frac{1}{4}x}$$
which is choice A.

Solve the Differential Equation 4y′=ex/4+y4 y ^ { \prime } = e ^ { x / 4 } + y4y′=ex/4+y

Solve the Differential Equation $4 y ^ { \prime } = e ^ { x / 4 } + y$