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)

(2 D^{2} - 3 D + 1) y = csc (2 π x)

d, d (x) = \frac{x ^{2} + 13 x + 42}{x ^{2} - 3 x - 28}

y^{^{''}} - 7 y^{^{'}} = x^{2}

x^{^{''}} + 3 x^{^{'}} + 2 x = 3 cos (2 t)

2 x^{^{''}} + 5 x^{^{'}} + 4 x = 10 r (t)

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)