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

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

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y^'=(2xy)/(x^2+1)

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

y = e^{c_{1}} x^{2} + e^{c_{1}}

Solution steps

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

Solve separable ODE:

y = e^{c_{1}} x^{2} + e^{c_{1}}

y = e^{c_{1}} x^{2} + e^{c_{1}}

f (x) = 6 + 1 - x, at x = - 8

\int \frac{x - 8}{( x + 4 ) ^{2} + 64} d x

\frac{d}{d t} (e^{- 4 t})

x \frac{d y}{d x} + 3 x^{2} + 4 y = 0

x \cdot e^{x}

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