Solve the differential equation.
-$$x y ^ { \prime } + 4 y = \frac { \cos x } { x ^ { 3 } } , x 0$$
A) $y = \frac { \cos x + C } { x ^ { 3 } } , x 0$
B) $y = \frac { \cos x + C } { x ^ { 4 } } , x 0$
C) $y = \frac { \sin x + C } { x ^ { 3 } } , x 0$
D) $y = \frac { \sin x + C } { x ^ { 4 } } , x 0$

Question

Solve the differential equation.
-$$x y ^ { \prime } + 4 y = \frac { \cos x } { x ^ { 3 } } , x > 0$$
A) $y = \frac { \cos x + C } { x ^ { 3 } } , x > 0$
B) $y = \frac { \cos x + C } { x ^ { 4 } } , x > 0$
C) $y = \frac { \sin x + C } { x ^ { 3 } } , x > 0$
D) $y = \frac { \sin x + C } { x ^ { 4 } } , x > 0$

Quizplus · Accepted Answer

The Answer of Solve the differential equation.
-$$x y ^ { \prime } + 4 y = \frac { \cos x } { x ^ { 3 } } , x > 0$$
A) $y = \frac { \cos x + C } { x ^ { 3 } } , x > 0$
B) $y = \frac { \cos x + C } { x ^ { 4 } } , x > 0$
C) $y = \frac { \sin x + C } { x ^ { 3 } } , x > 0$
D) $y = \frac { \sin x + C } { x ^ { 4 } } , x > 0$

Solve the Differential Equation xy′+4y=cos⁡xx3,x>0x y ^ { \prime } + 4 y = \frac { \cos x } { x ^ { 3 } } , x > 0xy′+4y=x3cosx​,x>0

Solve the Differential Equation $x y ^ { \prime } + 4 y = \frac { \cos x } { x ^ { 3 } } , x > 0$