What is the solution for y^{''}+y^'=4cos(2x) ?

The solution for y^{''}+y^'=4cos(2x) is y=c_{1}+c_{2}e^{-x}+2/5 sin(2x)-4/5 cos(2x)

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y^{''}+y^'=4cos(2x)

y^{^{''}} + y^{^{'}} = 4 cos (2 x)

y = c_{1} + c_{2} e^{- x} + \frac{2}{5} sin (2 x) - \frac{4}{5} cos (2 x)

Solution steps

y^{''} (x) + y^{'} (x) = 4 cos (2 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + \frac{2}{5} sin (2 x) - \frac{4}{5} cos (2 x)

y = c_{1} + c_{2} e^{- x} + \frac{2}{5} sin (2 x) - \frac{4}{5} cos (2 x)

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

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

y^{^{''}} + y^{^{'}} - 6 y = 5 x e^{3 x}

y^{^{''}} - 6 y + 9 y = t^{2} e^{3 t}

y^{^{''}} - 12 y^{^{'}} + 36 y = 14 e^{6 x}

What is the solution for y^{''}+y^'=4cos(2x) ?
The solution for y^{''}+y^'=4cos(2x) is y=c_{1}+c_{2}e^{-x}+2/5 sin(2x)-4/5 cos(2x)