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

The solution for y^{''}+4y^'=cos^2(x) is y=c_{1}+c_{2}e^{-4x}+(20x+8sin(2x)-4cos(2x)-5)/(160)

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

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

y = c_{1} + c_{2} e^{- 4 x} + \frac{20 x + 8 sin ( 2 x ) - 4 cos ( 2 x ) - 5}{160}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{20 x + 8 sin ( 2 x ) - 4 cos ( 2 x ) - 5}{160}

y = c_{1} + c_{2} e^{- 4 x} + \frac{20 x + 8 sin ( 2 x ) - 4 cos ( 2 x ) - 5}{160}

\int - \frac{1}{10 - x} d x

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

y = x^{2} - 8, y = 1

y = 5 e^{5 t + 1}

\frac{d}{d x} (ln (cos (x)) + x)

What is the solution for y^{''}+4y^'=cos^2(x) ?
The solution for y^{''}+4y^'=cos^2(x) is y=c_{1}+c_{2}e^{-4x}+(20x+8sin(2x)-4cos(2x)-5)/(160)