What is the solution for 3xdy+4ydx=x^2y^2dx ?

The solution for 3xdy+4ydx=x^2y^2dx is y= 2/(-x^2+2c_{1)x^{4/3}}

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3xdy+4ydx=x^2y^2dx

3 x d y + 4 y d x = x^{2} y^{2} d x

y = \frac{2}{- x ^{2} + 2 c _{1} x ^{\frac{4}{3}}}

Solution steps

3 x d y + 4 y d x = x^{2} y^{2} d x

Solve Bernoulli ODE:

y = \frac{2}{- x ^{2} + 2 c _{1} x ^{\frac{4}{3}}}

y = \frac{2}{- x ^{2} + 2 c _{1} x ^{\frac{4}{3}}}

\frac{d}{d x} (ln (18 x))

\int 2 x sec^{2} (4 x^{2}) d x

x \to (\frac{π}{2}) + lim (x - \frac{π}{2})

\frac{d}{d x} (4 x^{2} y + 2 z^{3})

x \to 0 lim ((tan (x))^{x})

What is the solution for 3xdy+4ydx=x^2y^2dx ?
The solution for 3xdy+4ydx=x^2y^2dx is y= 2/(-x^2+2c_{1)x^{4/3}}