What is the solution for y^{''}+y=sin(4x) ?

The solution for y^{''}+y=sin(4x) is y=c_{1}cos(x)+c_{2}sin(x)-1/15 sin(4x)

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y^{''}+y=sin(4x)

y^{^{''}} + y = sin (4 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{15} sin (4 x)

Solution steps

y^{''} (x) + y = sin (4 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{15} sin (4 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{15} sin (4 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = x e^{- 2 x}

\frac{d ^{2} x}{d t ^{2}} + 7 \frac{d x}{d t} + 6 x = 5 e^{- t}

y^{^{''}} + 2 y^{^{'}} + 5 = 2 cos (t + 0.523599)

y^{^{''}} + 4 y = cos (2 t), y (0) = 1, y^{^{'}} (0) = 1

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + x = 3 cos (t)

What is the solution for y^{''}+y=sin(4x) ?
The solution for y^{''}+y=sin(4x) is y=c_{1}cos(x)+c_{2}sin(x)-1/15 sin(4x)