What is the solution for (3x^2+2)^2y^'-xy^2=0,y(0)= 3/2 ?

The solution for (3x^2+2)^2y^'-xy^2=0,y(0)= 3/2 is y=(36x^2)/(21x^2+16)+(24)/(21x^2+16)

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(3x^2+2)^2y^'-xy^2=0,y(0)= 3/2

(3 x^{2} + 2)^{2} y^{^{'}} - x y^{2} = 0, y (0) = \frac{3}{2}

y = \frac{36 x ^{2}}{21 x ^{2} + 16} + \frac{24}{21 x ^{2} + 16}

Solution steps

(3 x^{2} + 2)^{2} y^{'} (x) - x y^{2} = 0

Solve separable ODE:

y = \frac{36 x ^{2}}{21 x ^{2} + 16} + \frac{24}{21 x ^{2} + 16}

y = \frac{36 x ^{2}}{21 x ^{2} + 16} + \frac{24}{21 x ^{2} + 16}

\frac{x d x}{1 + x ^{2}} = \frac{d y}{y}

\frac{d y}{d x} = 2 x y, y (1) = 5

x^{^{'}} = 20 t - \frac{4 x}{30} t

2 x y \frac{d y}{d x} = 6 + 14 ln (x)

(1 + x^{2}) \frac{d y}{d x} = sin (y)

What is the solution for (3x^2+2)^2y^'-xy^2=0,y(0)= 3/2 ?
The solution for (3x^2+2)^2y^'-xy^2=0,y(0)= 3/2 is y=(36x^2)/(21x^2+16)+(24)/(21x^2+16)