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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(5 y - 2 x) d y - 2 y d x = 0

\frac{\partial}{\partial x} (\frac{3 x - 5 y}{3 x + 5 y})

\int sin (8 x) d x

\frac{\partial}{\partial y} (e^{x y} + 2 x - 5 y)

y = 2 x^{3} - 5 x + 1, (1, - 2)

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