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^3)/3+(e^6x^2)/2+(e^6x)/4

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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 ^{3}}{3} + \frac{e ^{6} x ^{2}}{2} + \frac{e ^{6} x}{4}

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 ^{3}}{3} + \frac{e ^{6} x ^{2}}{2} + \frac{e ^{6} x}{4}

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

y^{^{''}} + 64 y = sin (8 x)

y^{^{''}} + 6 y = t^{4}, y (0) = 0, y^{^{'}} (0) = 0

(D^{2} + 2 D - 8) y = (6 x^{- 1} - x^{- 2}) e^{2 x}

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

y^{^{''}} - y = 5, e^{- 4 x} + 2 x y (0) = 0

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^3)/3+(e^6x^2)/2+(e^6x)/4