What is the solution for y^{''}-8y^'+16y=xe^{4x} ?

The solution for y^{''}-8y^'+16y=xe^{4x} is y=c_{1}e^{4x}+c_{2}xe^{4x}+(e^{4x}x^3)/6

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y^{''}-8y^'+16y=xe^{4x}

y^{^{''}} - 8 y^{^{'}} + 16 y = x e^{4 x}

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

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = x e^{4 x}

Solve linear ODE:

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

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

(d^{2} - 4 d + 3) y = 2 x e^{3 x} + 3 e^{x} cos (2 x)

x^{^{''}} + 6 x^{^{'}} + 23 x = 8 cos (4 t)

y^{^{''}} - 7 y^{^{'}} + 12 y = 4 e^{x}

y^{^{''}} + 3 y = x^{2} sin (x)

4 y^{^{''}} + 36 y = csc (x)

What is the solution for y^{''}-8y^'+16y=xe^{4x} ?
The solution for y^{''}-8y^'+16y=xe^{4x} is y=c_{1}e^{4x}+c_{2}xe^{4x}+(e^{4x}x^3)/6