Solve the differential equation.
-$$x \frac { d y } { d x } + 2 y = 4 - \frac { 1 } { x } , x 0$$
A) $y = \frac { 4 x ^ { 2 } - 2 x + C } { 2 x ^ { 2 } } , x 0$
B) $y = \frac { 4 x ^ { 2 } - x + C } { 2 x ^ { 2 } } , x 0$
C) $y = \frac { x ^ { 2 } - 4 x + C } { 2 x ^ { 2 } } , x 0$
D) $y = \frac { 4 x ^ { 2 } + 2 x + C } { 2 x ^ { 2 } } , x 0$

Question

Solve the differential equation.
-$$x \frac { d y } { d x } + 2 y = 4 - \frac { 1 } { x } , x > 0$$
A) $y = \frac { 4 x ^ { 2 } - 2 x + C } { 2 x ^ { 2 } } , x > 0$
B) $y = \frac { 4 x ^ { 2 } - x + C } { 2 x ^ { 2 } } , x > 0$
C) $y = \frac { x ^ { 2 } - 4 x + C } { 2 x ^ { 2 } } , x > 0$
D) $y = \frac { 4 x ^ { 2 } + 2 x + C } { 2 x ^ { 2 } } , x > 0$

Quizplus · Accepted Answer

This differential equation can be solved by making it a linear first-order differential equation in the form $x\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = 2$ and $Q(x) = 4 - \frac{1}{x}$. The integrating factor is $e^{\int P(x)dx} = e^{\int 2dx} = e^{2\ln|x|} = x^2$. Multiplying both sides by $x^2$, we get $x^3\frac{dy}{dx} + 2x^2y = 4x^2 - x$. Integrating both sides with respect to $x$, we find $y = \frac{4x^2 - 2x + C}{2x^2}$, where $C$ is the constant of integration.

Solve the Differential Equation xdydx+2y=4−1x,x>0x \frac { d y } { d x } + 2 y = 4 - \frac { 1 } { x } , x > 0xdxdy​+2y=4−x1​,x>0

Solve the Differential Equation $x \frac { d y } { d x } + 2 y = 4 - \frac { 1 } { x } , x > 0$