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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{3 e ^{- 4 x}}{x}

(d^{3} - 3 d^{2} - 10 d + 24) y = 2 x^{2} + 5

y^{^{''}} - y^{^{'}} + 2 = 0

y^{^{'''}} - y^{^{'}} = x e^{(- x)} + 5 cos (x)

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

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