What is the solution for y^{''}-4y^'=8x^2+2e^{3x} ?

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

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y^{''}-4y^'=8x^2+2e^{3x}

y^{^{''}} - 4 y^{^{'}} = 8 x^{2} + 2 e^{3 x}

y = c_{1} e^{4 x} + c_{2} - \frac{2 x ^{3}}{3} - \frac{x ^{2}}{2} - \frac{x}{4} - \frac{2}{3} e^{3 x}

Solution steps

y^{''} (x) - 4 y^{'} (x) = 8 x^{2} + 2 e^{3 x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} - \frac{2 x ^{3}}{3} - \frac{x ^{2}}{2} - \frac{x}{4} - \frac{2}{3} e^{3 x}

y = c_{1} e^{4 x} + c_{2} - \frac{2 x ^{3}}{3} - \frac{x ^{2}}{2} - \frac{x}{4} - \frac{2}{3} e^{3 x}

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

x^{^{''}} + 7 x^{^{'}} + 10 x = 20, x (0) = 5, x^{^{'}} (0) = 3

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 x^{2} + 1

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

\frac{d ^{2} x}{d t ^{2}} + 16 x = 1 + 2 sin (4 t)

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