What is the solution for y^{''}+y= 1/4 cos(3t) ?

The solution for y^{''}+y= 1/4 cos(3t) is y=c_{1}cos(t)+c_{2}sin(t)-1/32 cos(3t)

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y^{''}+y= 1/4 cos(3t)

y^{^{''}} + y = \frac{1}{4} cos (3 t)

y = c_{1} cos (t) + c_{2} sin (t) - \frac{1}{32} cos (3 t)

Solution steps

y^{''} (t) + y = \frac{1}{4} cos (3 t)

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) - \frac{1}{32} cos (3 t)

y = c_{1} cos (t) + c_{2} sin (t) - \frac{1}{32} cos (3 t)

\int_{1}^{4} 16 x^{5} d x

\frac{d y}{d x} = - 2 x (y - 3)

y^{^{''}} - 4 y^{^{'}} + 3 y = x, y (0) = 1, y^{^{'}} (0) = 0

\int_{1}^{e} \frac{8}{z}

What is the solution for y^{''}+y= 1/4 cos(3t) ?
The solution for y^{''}+y= 1/4 cos(3t) is y=c_{1}cos(t)+c_{2}sin(t)-1/32 cos(3t)