What is the solution for y^{''}-4y^'=x*e^{4x} ?

The solution for y^{''}-4y^'=x*e^{4x} is y=c_{1}e^{4x}+c_{2}+(e^{4x}x^2)/8-1/64 (4e^{4x}x-e^{4x})

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y^{''}-4y^'=x*e^{4x}

y^{^{''}} - 4 y^{^{'}} = x \cdot e^{4 x}

y = c_{1} e^{4 x} + c_{2} + \frac{e ^{4 x} x ^{2}}{8} - \frac{1}{64} (4 e^{4 x} x - e^{4 x})

Solution steps

y^{''} (x) - 4 y^{'} (x) = x \cdot e^{4 x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} + \frac{e ^{4 x} x ^{2}}{8} - \frac{1}{64} (4 e^{4 x} x - e^{4 x})

y = c_{1} e^{4 x} + c_{2} + \frac{e ^{4 x} x ^{2}}{8} - \frac{1}{64} (4 e^{4 x} x - e^{4 x})

y^{^{''}} + 2 y^{^{'}} + y = e^{- t} (t^{3} - 2 t)

(d^{2} - 6 d + 3) y = e^{3} x

y^{^{''}} + 4 y^{^{'}} + 3 y = 6

y^{^{''}} + 4 y = 1, 6 t y (0) = 3, y^{^{'}} (0) = - 6

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

What is the solution for y^{''}-4y^'=x*e^{4x} ?
The solution for y^{''}-4y^'=x*e^{4x} is y=c_{1}e^{4x}+c_{2}+(e^{4x}x^2)/8-1/64 (4e^{4x}x-e^{4x})