What is the solution for y^{''}+x+4y=0 ?

The solution for y^{''}+x+4y=0 is y=c_{1}cos(2x)+c_{2}sin(2x)-x/4

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y^{''}+x+4y=0

y^{^{''}} + x + 4 y = 0

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x}{4}

Solution steps

y^{''} (x) + x + 4 y = 0

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x}{4}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{x}{4}

y^{^{''}} + 6 y^{^{'}} + 9 y = x e^{4 x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 3

y^{^{''}} + 3 y^{^{'}} + 2 y = 2

y^{^{''}} + 2 y^{^{'}} + y = ln (x) \cdot e^{- x}

x^{^{''}} - 14 x^{^{'}} + 49 x = 5 t e^{7 t}

What is the solution for y^{''}+x+4y=0 ?
The solution for y^{''}+x+4y=0 is y=c_{1}cos(2x)+c_{2}sin(2x)-x/4