What is the solution for x^2y^'+2x^3y=y^2(1+2x^2) ?

The solution for x^2y^'+2x^3y=y^2(1+2x^2) is y= x/(1+c_{1)e^{x^2}x}

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x^2y^'+2x^3y=y^2(1+2x^2)

x^{2} y^{^{'}} + 2 x^{3} y = y^{2} (1 + 2 x^{2})

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

Solution steps

x^{2} y^{'} (x) + 2 x^{3} y = y^{2} (1 + 2 x^{2})

Solve Bernoulli ODE:

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

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

\int sin (10 x) cos (3 x) d x

\frac{d}{d x} (9 x + 2)

\int \frac{x + 2 x - 1}{2 x x - 1} d x

\frac{( x ^{3} - 3 )}{( x ^{2} + x - 1 )}

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

What is the solution for x^2y^'+2x^3y=y^2(1+2x^2) ?
The solution for x^2y^'+2x^3y=y^2(1+2x^2) is y= x/(1+c_{1)e^{x^2}x}