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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} - 25 x = e^{5 t}

y^{^{'''}} + y^{^{''}} - 2 y = - e^{3 x} (9 + 67 x + 17 x^{2})

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = x + 6 e^{- x}

3 x^{^{''}} + 48 x = 5 cos (4 t)

8 y^{^{''}} + 6 y^{^{'}} - y = 11

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