What is the solution for x(dy)/(dx)-3y=y^2,y(1)=3 ?

The solution for x(dy)/(dx)-3y=y^2,y(1)=3 is y=(3x^3)/(2-x^3)

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x(dy)/(dx)-3y=y^2,y(1)=3

x \frac{d y}{d x} - 3 y = y^{2}, y (1) = 3

y = \frac{3 x ^{3}}{2 - x ^{3}}

Solution steps

x \frac{d y}{d x} - 3 y = y^{2}

Solve Bernoulli ODE:

y = \frac{3 x ^{3}}{2 - x ^{3}}

y = \frac{3 x ^{3}}{2 - x ^{3}}

\int \frac{- y}{x ^{2} + y ^{2}} d x

\int 5 cos (x) + 2 sin (x) d x

f (x) = 3 x^{\frac{2}{3}} - 3, at x = 27

\frac{d}{d x} (12 x^{\frac{1}{2}})

\int sin (9)

What is the solution for x(dy)/(dx)-3y=y^2,y(1)=3 ?
The solution for x(dy)/(dx)-3y=y^2,y(1)=3 is y=(3x^3)/(2-x^3)