What is the solution for y^{''}+4y=cos(4t) ?

The solution for y^{''}+4y=cos(4t) is y=c_{1}cos(2t)+c_{2}sin(2t)-1/12 cos(4t)

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y^{''}+4y=cos(4t)

y^{^{''}} + 4 y = cos (4 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{12} cos (4 t)

Solution steps

y^{''} (t) + 4 y = cos (4 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{12} cos (4 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{12} cos (4 t)

y^{^{''}} - y^{^{'}} - 2 y = x e^{x}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 = e^{- 5 x}

(d^{2} - 9) y = 30 sin^{3} (x)

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 6

y^{^{''}} + 4 y^{^{'}} + 5 y = 13 sin (2 t)

What is the solution for y^{''}+4y=cos(4t) ?
The solution for y^{''}+4y=cos(4t) is y=c_{1}cos(2t)+c_{2}sin(2t)-1/12 cos(4t)