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

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

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y^{''}+k^2y=k_{2}*sin(kx)

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

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

Solution steps

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

Solve linear ODE:

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

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

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