What is the solution for y^{''}-4y^'+4y=(18x-4)e^{2x} ?

The solution for y^{''}-4y^'+4y=(18x-4)e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+3x^3e^{2x}-2e^{2x}x^2

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y^{''}-4y^'+4y=(18x-4)e^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = (18 x - 4) e^{2 x}

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = (18 x - 4) e^{2 x}

Solve linear ODE:

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

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

\frac{d y}{d t} - y cos (t) = cos (t)

\int_{0}^{l n (6)} e^{x} - e^{- 2 x} d x

\int x - 2 d x

\int 6 sin^{3} (x) cos^{2} (x) d x

\int x cos^{2} (x) sin (x) d x

What is the solution for y^{''}-4y^'+4y=(18x-4)e^{2x} ?
The solution for y^{''}-4y^'+4y=(18x-4)e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+3x^3e^{2x}-2e^{2x}x^2