What is the solution for y^{''}-4y=3xe^{2x} ?

The solution for y^{''}-4y=3xe^{2x} is y=c_{1}e^{2x}+c_{2}e^{-2x}+(3e^{2x}x^2)/8-(3e^{2x}x)/(16)

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y^{''}-4y=3xe^{2x}

y^{^{''}} - 4 y = 3 x e^{2 x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{3 e ^{2 x} x ^{2}}{8} - \frac{3 e ^{2 x} x}{16}

Solution steps

y^{''} (x) - 4 y = 3 x e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{3 e ^{2 x} x ^{2}}{8} - \frac{3 e ^{2 x} x}{16}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{3 e ^{2 x} x ^{2}}{8} - \frac{3 e ^{2 x} x}{16}

y^{^{''}} - 25 y = 3 x^{2} - 2 + e^{x}

y^{^{'''}} - y^{^{''}} = x

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{x} - 12

y^{^{''}} - 2 y^{^{'}} + 2 y = cos (t)

What is the solution for y^{''}-4y=3xe^{2x} ?
The solution for y^{''}-4y=3xe^{2x} is y=c_{1}e^{2x}+c_{2}e^{-2x}+(3e^{2x}x^2)/8-(3e^{2x}x)/(16)