What is the solution for y^{''}-2y^'+5y=2(cos^2(x))e^x ?

The solution for y^{''}-2y^'+5y=2(cos^2(x))e^x is

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y^{''}-2y^'+5y=2(cos^2(x))e^x

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 (cos^{2} (x)) e^{x}

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + 5 y = 2 (cos^{2} (x)) e^{x}

Solve linear ODE:

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

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

x^{^{''}} - 4 x^{^{'}} + 4 x = 4 e^{2 t}, x^{0} = - 1, x^{^{'}} 0 = - 4

y^{^{''}} - 12 y^{^{'}} + 36 y = 10 e^{6 x}

y^{^{''}} - 2 y^{^{'}} + y = 14 x^{\frac{2}{3}} e^{x}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = 4 e^{2 x}

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

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