What is the solution for xydx+(10xy+x)dy=0 ?

The solution for xydx+(10xy+x)dy=0 is y=(\W(\frac{10)/(e^{x-c_{1)}})}{10}

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xydx+(10xy+x)dy=0

x y d x + (10 x y + x) d y = 0

y = \frac{W ( \frac{10}{e ^{x - c_{1}}} )}{10}

Solution steps

x y d x + (10 x y + x) d y = 0

Solve separable ODE:

y = \frac{W ( \frac{10}{e ^{x - c_{1}}} )}{10}

y = \frac{W ( \frac{10}{e ^{x - c_{1}}} )}{10}

x \frac{d y}{d x} = y tan (ln (y))

\frac{d y}{d t} = y^{2} (8 + t)

y^{^{'}} y = y^{2}

\frac{d y}{d x} = \frac{x y ^{2}}{x}

(2 + cos (t)) (\frac{d y}{d t}) - y sin (t) = 0

What is the solution for xydx+(10xy+x)dy=0 ?
The solution for xydx+(10xy+x)dy=0 is y=(\W(\frac{10)/(e^{x-c_{1)}})}{10}