What is the solution for y^{''}+4y=r(x) ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 5 x

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{x} + e^{x} x^{2} + 5 cos (2 x)

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x - 2, y^{^{'}} (π) = π, y (0) = 0

\frac{d ^{2} y}{d x ^{2}} + 4 y = sin^{2} (x)

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

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