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

The solution for y^{''}-4y^'=8xe^{3x} is y=c_{1}e^{4x}+c_{2}-(8e^{3x}x)/3-16/9 e^{3x}

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y^{''}-4y^'=8xe^{3x}

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

y = c_{1} e^{4 x} + c_{2} - \frac{8 e ^{3 x} x}{3} - \frac{16}{9} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} - \frac{8 e ^{3 x} x}{3} - \frac{16}{9} e^{3 x}

y = c_{1} e^{4 x} + c_{2} - \frac{8 e ^{3 x} x}{3} - \frac{16}{9} e^{3 x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{x}

y^{^{''}} - 4 y^{^{'}} + 7 y = t e^{t}

y^{^{''}} + 9 y = sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y = 8 sin (2 x) + 5 e^{x}

y^{^{''}} - y^{^{'}} - 6 y = - e^{x} + 12 x, y (0) = 1, y^{^{'}} (0) = - 2

What is the solution for y^{''}-4y^'=8xe^{3x} ?
The solution for y^{''}-4y^'=8xe^{3x} is y=c_{1}e^{4x}+c_{2}-(8e^{3x}x)/3-16/9 e^{3x}