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}}

y^{^{''}} + y = cosh (x)

(D^{2} + D - 2) y = 2 x - 40 cos (2 x)

y^{^{''''}} + 16 y^{^{''}} = sin (4 t) + t e^{5 t} + 4

y^{^{''}} - 4 y^{^{'}} + 3 y = e^{3 x}

y^{^{''}} + 9 y = t^{2} e^{3 t} + 6

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)