What is the solution for y^{''}-4y^'=8e^{6x}(x-1) ?

The solution for y^{''}-4y^'=8e^{6x}(x-1) is y=c_{1}e^{4x}+c_{2}+(2e^{6x}x)/3-10/9 e^{6x}

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y^{''}-4y^'=8e^{6x}(x-1)

y^{^{''}} - 4 y^{^{'}} = 8 e^{6 x} (x - 1)

y = c_{1} e^{4 x} + c_{2} + \frac{2 e ^{6 x} x}{3} - \frac{10}{9} e^{6 x}

Solution steps

y^{''} (x) - 4 y^{'} (x) = 8 e^{6 x} (x - 1)

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} + \frac{2 e ^{6 x} x}{3} - \frac{10}{9} e^{6 x}

y = c_{1} e^{4 x} + c_{2} + \frac{2 e ^{6 x} x}{3} - \frac{10}{9} e^{6 x}

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

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

\frac{d ^{2} y}{d t ^{2}} = 0.5 t - y

y^{^{''}} - y = e^{x} cos (x)

y^{^{'''}} - y^{^{'}} = 7 t

What is the solution for y^{''}-4y^'=8e^{6x}(x-1) ?
The solution for y^{''}-4y^'=8e^{6x}(x-1) is y=c_{1}e^{4x}+c_{2}+(2e^{6x}x)/3-10/9 e^{6x}