What is the solution for y^{''}+16y=sin(3x) ?

The solution for y^{''}+16y=sin(3x) is y=c_{1}cos(4x)+c_{2}sin(4x)+1/7 sin(3x)

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y^{''}+16y=sin(3x)

y^{^{''}} + 16 y = sin (3 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{7} sin (3 x)

Solution steps

y^{''} (x) + 16 y = sin (3 x)

Solve linear ODE:

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{7} sin (3 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) + \frac{1}{7} sin (3 x)

2 x^{^{''}} - x^{^{'}} + 3 t = 5

y^{^{''}} - 2 y^{^{'}} + 82 y = e x cos (9 x)

y^{^{''}} - 9 y^{^{'}} + 6 y = x e^{x}

y^{^{''}} - 2 y^{^{'}} + 2 y = 5 x^{2} - 9 x + 5

y^{^{''}} + 5 y = - 5 - 6 x + 12 x^{2} - 5 x^{3} + 5 x^{4}

What is the solution for y^{''}+16y=sin(3x) ?
The solution for y^{''}+16y=sin(3x) is y=c_{1}cos(4x)+c_{2}sin(4x)+1/7 sin(3x)