What is the solution for xln(x)(dy)/(dx)+y=xe^x ?

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

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xln(x)(dy)/(dx)+y=xe^x

x ln (x) \frac{d y}{d x} + y = x e^{x}

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

Solution steps

x ln (x) \frac{d y}{d x} + y = x e^{x}

Solve linear ODE:

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

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

\int ∣ x - 1 ∣ d x

\int \frac{- 2}{x ^{2} + 16} d x

e^{\frac{y}{x}} \frac{( - y d x + x d y )}{x ^{2}} = 0

f (x) = \frac{8 x}{x ^{2} - 1}, (3, 3)

\frac{d}{d x} (3 x^{2} - 5 x + 4)

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