What is the solution for y^'=((y^2+2xy))/(x^2) ?

The solution for y^'=((y^2+2xy))/(x^2) is y=(x^2)/(-x+c_{1)}

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y^'=((y^2+2xy))/(x^2)

y^{^{'}} = \frac{( y ^{2} + 2 x y )}{x ^{2}}

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

Solution steps

y^{'} (x) = \frac{( y ^{2} + 2 x y )}{x ^{2}}

Solve Bernoulli ODE:

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

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

\int e^{3 x} sin (3 x) d x

y^{^{'}} = \frac{x}{y}, y (0) = 3

\frac{d}{d x} (\frac{18 x + 4 x ^{2}}{( 9 + 4 x ) ^{2}})

\int sin (t^{2}) d t

l a pl a ce t r an s f or m 1 - cos (2 t)

What is the solution for y^'=((y^2+2xy))/(x^2) ?
The solution for y^'=((y^2+2xy))/(x^2) is y=(x^2)/(-x+c_{1)}