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

The solution for y^{''}-4y^'=2cos^2(4x) is y=c_{1}e^{4x}+c_{2}+(-40x-2cos(8x)-sin(8x)-10)/(160)

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

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

y = c_{1} e^{4 x} + c_{2} + \frac{- 40 x - 2 cos ( 8 x ) - sin ( 8 x ) - 10}{160}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} + \frac{- 40 x - 2 cos ( 8 x ) - sin ( 8 x ) - 10}{160}

y = c_{1} e^{4 x} + c_{2} + \frac{- 40 x - 2 cos ( 8 x ) - sin ( 8 x ) - 10}{160}

y^{^{''}} + 2 y^{^{'}} - 3 y = 16 e^{- 3 t}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{- t}

y^{^{''}} - y^{^{'}} + 196 y = 14 sin (14 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 6 sin (2 x) + 7 cos (2 x)

(D^{2} - 3 D + 2) y = 2 x^{3} - 9 x^{2} + 6 x

What is the solution for y^{''}-4y^'=2cos^2(4x) ?
The solution for y^{''}-4y^'=2cos^2(4x) is y=c_{1}e^{4x}+c_{2}+(-40x-2cos(8x)-sin(8x)-10)/(160)