What is the solution for y^{''}-4y^'+4y=8x^2+e^{2x} ?

The solution for y^{''}-4y^'+4y=8x^2+e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+2x^2+4x+3+(e^{2x}x^2)/2

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y^{''}-4y^'+4y=8x^2+e^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 8 x^{2} + e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 2 x^{2} + 4 x + 3 + \frac{e ^{2 x} x ^{2}}{2}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 8 x^{2} + e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 2 x^{2} + 4 x + 3 + \frac{e ^{2 x} x ^{2}}{2}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 2 x^{2} + 4 x + 3 + \frac{e ^{2 x} x ^{2}}{2}

What is the solution for y^{''}-4y^'+4y=8x^2+e^{2x} ?
The solution for y^{''}-4y^'+4y=8x^2+e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+2x^2+4x+3+(e^{2x}x^2)/2