What is the solution for y^{''}-4y^'+4y=5e^{2x} ?

The solution for y^{''}-4y^'+4y=5e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+(5e^{2x}x^2)/2

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y^{''}-4y^'+4y=5e^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 5 e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{5 e ^{2 x} x ^{2}}{2}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 5 e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{5 e ^{2 x} x ^{2}}{2}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{5 e ^{2 x} x ^{2}}{2}

y^{^{''}} + 3 y = 2 x

y^{^{''}} + 16 y = cos (4 t), y (0) = 0, y^{^{'}} (0) = 1

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

y^{^{''}} + 2 y^{^{'}} - 15 y = - 53 t + 30 t^{2}

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{x}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-4y^'+4y=5e^{2x} ?
The solution for y^{''}-4y^'+4y=5e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+(5e^{2x}x^2)/2