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

The solution for y^{''}+y^'= 1/4 cos(3t) is y=c_{1}+c_{2}e^{-t}+1/120 sin(3t)-1/40 cos(3t)

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

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

y = c_{1} + c_{2} e^{- t} + \frac{1}{120} sin (3 t) - \frac{1}{40} cos (3 t)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + \frac{1}{120} sin (3 t) - \frac{1}{40} cos (3 t)

y = c_{1} + c_{2} e^{- t} + \frac{1}{120} sin (3 t) - \frac{1}{40} cos (3 t)

\frac{\partial}{\partial x} (e^{2 y - x})

\frac{x ^{2}}{x - 1}

\int 5 x^{2} sin (2 x) d x

\int cos^{3} (5 x) d x

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

What is the solution for y^{''}+y^'= 1/4 cos(3t) ?
The solution for y^{''}+y^'= 1/4 cos(3t) is y=c_{1}+c_{2}e^{-t}+1/120 sin(3t)-1/40 cos(3t)