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

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

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{2 x} x^{2} + \frac{x + 1}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{2 x} x^{2} + \frac{x + 1}{8}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{2 x} x^{2} + \frac{x + 1}{8}

y^{^{''}} - y^{^{'}} = t^{2}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 x

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 7 e^{- t} + 2

y^{^{'''}} + y^{^{''}} = e^{t} + t + 1

y^{^{''}} - y^{^{'}} = - 4

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