What is the solution for y^{''}+k^2y=ksin(kx) ?

The solution for y^{''}+k^2y=ksin(kx) is y=c_{1}cos(kx)+c_{2}sin(kx)-(xcos(kx))/2

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y^{''}+k^2y=ksin(kx)

y^{^{''}} + k^{2} y = k sin (k x)

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

Solution steps

y^{''} (x) + k^{2} y = k sin (k x)

Solve linear ODE:

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

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

2 y^{^{''}} + y^{^{'}} + 1 = 2 e^{2} x

y^{^{''}} + 2 y^{^{'}} + y = 1 + x e^{- x}

y^{^{'''}} - 2 y^{^{'}} = 2, y^{^{''}} (0) = - 1

y^{^{''}} - y = 18 e^{5 x}

y^{^{''}} + 4 y^{^{'}} + 8 y = 160

What is the solution for y^{''}+k^2y=ksin(kx) ?
The solution for y^{''}+k^2y=ksin(kx) is y=c_{1}cos(kx)+c_{2}sin(kx)-(xcos(kx))/2