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

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

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

8 y^{^{''}} + 4 y^{^{'}} + 5 y = 1, y (0) = 1

y^{^{'''}} - 4 y^{^{'}} = cot (2 x)

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

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

y^{^{''}} + 3 y^{^{'}} - 10 y = e^{x} cos (3 x)

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