What is the solution for xy^'+y+yln(xy)=0 ?

The solution for xy^'+y+yln(xy)=0 is y=(e^{(c_{1})/x})/(x)

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xy^'+y+yln(xy)=0

x y^{^{'}} + y + y ln (x y) = 0

y = \frac{e ^{\frac{c _{1}}{x}}}{x}

Solution steps

x y^{'} (x) + y + y ln (x y) = 0

Solve using substitution:

y = \frac{e ^{\frac{c _{1}}{x}}}{x}

y = \frac{e ^{\frac{c _{1}}{x}}}{x}

r \to 0 lim (\frac{1}{r ^{2}})

\int 6 e^{- 0.4 x} d x

\int \frac{6}{2} d x

\frac{d}{d x} (x^{2} \cdot sin^{2} (x))

f (x) = \frac{36 x}{x ^{2} + 36}, (- 3, - \frac{12}{5})

What is the solution for xy^'+y+yln(xy)=0 ?
The solution for xy^'+y+yln(xy)=0 is y=(e^{(c_{1})/x})/(x)