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}

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)