What is the solution for xln(x)y^'+y=8xe^x ?

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

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xln(x)y^'+y=8xe^x

x ln (x) y^{^{'}} + y = 8 x e^{x}

y = \frac{8 e ^{x}}{ln ( x )} + \frac{c _{1}}{ln ( x )}

Solution steps

x ln (x) y^{'} (x) + y = 8 x e^{x}

Solve linear ODE:

y = \frac{8 e ^{x}}{ln ( x )} + \frac{c _{1}}{ln ( x )}

y = \frac{8 e ^{x}}{ln ( x )} + \frac{c _{1}}{ln ( x )}

\frac{x}{1 - x ^{2}}

n = 0 \sum \infty \frac{1}{n ln ( n )}

\frac{\partial}{\partial x} (x^{2} y + 2 y^{2} - 4 x y + 6)

\int \frac{y ^{2} + 20 y + 100}{( y ^{2} + 100 ) ^{2}} d y

y^{^{''}} - 8 y^{^{'}} + 19 y = 0

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