What is the solution for y^{''}-4y^'+4y=(x+1)e^x ?

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

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y^{''}-4y^'+4y=(x+1)e^x

y^{^{''}} - 4 y^{^{'}} + 4 y = (x + 1) e^{x}

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = (x + 1) e^{x}

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} = e^{2} x, y (0) = 0, y (1) = 0

y^{^{''}} - 4 y = sin (x)

(D^{2} + D) y = - cos (x)

4 y^{^{''}} + 2 y^{^{'}} + 70 y = 4 cos (4 t)

y^{^{''''}} - 2 y^{^{''}} - 8 y = 6 e^{2 x}

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