What is the solution for 2ydx+x(x^2ln(y)-1)dy=0 ?

The solution for 2ydx+x(x^2ln(y)-1)dy=0 is y=-(c_{1})/(\W(-\frac{c_{1)}{e^{(x^2+1)/(x^2)}})}

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2ydx+x(x^2ln(y)-1)dy=0

2 y d x + x (x^{2} ln (y) - 1) d y = 0

y = - \frac{c _{1}}{W ( - \frac{c _{1}}{e ^{\frac{x ^{2} + 1}{x ^{2}}}} )}

Solution steps

2 y d x + x (x^{2} ln (y) - 1) d y = 0

Solve Bernoulli ODE:

y = - \frac{c _{1}}{W ( - \frac{c _{1}}{e ^{\frac{x ^{2} + 1}{x ^{2}}}} )}

y = - \frac{c _{1}}{W ( - \frac{c _{1}}{e ^{\frac{x ^{2} + 1}{x ^{2}}}} )}

\frac{d}{d x} (sec (5 x + 2))

y = \frac{1}{2} x^{3} + 2, y = x + 1, x = 0, x = 1

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

\frac{\partial}{\partial x} (e^{x^{2} + y^{2} + 2 y + 5})

x \to 2 lim (\frac{- 2 x ^{2}}{x - 2})

What is the solution for 2ydx+x(x^2ln(y)-1)dy=0 ?
The solution for 2ydx+x(x^2ln(y)-1)dy=0 is y=-(c_{1})/(\W(-\frac{c_{1)}{e^{(x^2+1)/(x^2)}})}