What is the solution for y^'= 3/x (y^2+y),y(4)=2 ?

The solution for y^'= 3/x (y^2+y),y(4)=2 is y=(x^3)/(96-x^3)

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y^'= 3/x (y^2+y),y(4)=2

y^{^{'}} = \frac{3}{x} (y^{2} + y), y (4) = 2

y = \frac{x ^{3}}{96 - x ^{3}}

Solution steps

y^{'} (x) = \frac{3}{x} (y^{2} + y)

Solve separable ODE:

y = \frac{x ^{3}}{96 - x ^{3}}

y = \frac{x ^{3}}{96 - x ^{3}}

\int \frac{1}{v} d v

5 \cdot e^{- 3 x} \cdot sin (5 x) \cdot x

x \to 3 lim (5 - x)

\frac{1}{16} tan^{2} (x) + cos (x)

x \to 1 + lim (3 x)

What is the solution for y^'= 3/x (y^2+y),y(4)=2 ?
The solution for y^'= 3/x (y^2+y),y(4)=2 is y=(x^3)/(96-x^3)