What is the solution for x(dy)/(dx)+yln(x)-y=yln(y) ?

The solution for x(dy)/(dx)+yln(x)-y=yln(y) is y=e^{c_{1}x}x

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x(dy)/(dx)+yln(x)-y=yln(y)

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

y = e^{c_{1} x} x

Solution steps

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

Solve using substitution:

y = e^{c_{1} x} x

y = e^{c_{1} x} x

\frac{\partial}{\partial y} (\frac{1}{x ^{2} yz})

\frac{d y}{d t} - \frac{5}{t} y = t^{5}

\frac{d}{d x} (\frac{- 1 x + 1 + x}{x ^{2} - 1})

\frac{d}{d x} (1 - 5 cos (x))

x \to + 3 lim (6 x^{2})

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