What is the solution for y^{''}-8y^'+16y=(e^{4x})/x ?

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

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

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

y = c_{1} e^{4 x} + c_{2} x e^{4 x} - e^{4 x} x + e^{4 x} x ln (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} x e^{4 x} - e^{4 x} x + e^{4 x} x ln (x)

y = c_{1} e^{4 x} + c_{2} x e^{4 x} - e^{4 x} x + e^{4 x} x ln (x)

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