What is the solution for y^{''}-4y^'+8y=xe^x ?

The solution for y^{''}-4y^'+8y=xe^x is y=e^{2x}(c_{1}cos(2x)+c_{2}sin(2x))+(e^xx)/5+2/25 e^x

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y^{''}-4y^'+8y=xe^x

y^{^{''}} - 4 y^{^{'}} + 8 y = x e^{x}

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{e ^{x} x}{5} + \frac{2}{25} e^{x}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 8 y = x e^{x}

Solve linear ODE:

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{e ^{x} x}{5} + \frac{2}{25} e^{x}

y = e^{2 x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{e ^{x} x}{5} + \frac{2}{25} e^{x}

What is the solution for y^{''}-4y^'+8y=xe^x ?
The solution for y^{''}-4y^'+8y=xe^x is y=e^{2x}(c_{1}cos(2x)+c_{2}sin(2x))+(e^xx)/5+2/25 e^x