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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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