What is the solution for y^{''}-4y^'+4y=120x^2e^{2x} ?

The solution for y^{''}-4y^'+4y=120x^2e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+10e^{2x}x^4

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y^{''}-4y^'+4y=120x^2e^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 120 x^{2} e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 10 e^{2 x} x^{4}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 120 x^{2} e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 10 e^{2 x} x^{4}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 10 e^{2 x} x^{4}

x \to - 4 lim (\frac{x ^{2} + 5}{x + 4})

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

\frac{d}{d z} (- z)

\int \frac{1}{x ^{2} + 4 x + 6} d x

ln (x + 6)

What is the solution for y^{''}-4y^'+4y=120x^2e^{2x} ?
The solution for y^{''}-4y^'+4y=120x^2e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+10e^{2x}x^4