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

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

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

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

y = e^{c_{1}} x^{2} + 3 e^{c_{1}} + 1

Solution steps

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

Solve separable ODE:

y = e^{c_{1}} x^{2} + 3 e^{c_{1}} + 1

y = e^{c_{1}} x^{2} + 3 e^{c_{1}} + 1

\int 19 sin^{3} (x) cos^{2} (x) d x

f (x) = \frac{6}{x ^{2}}, at x = 1

y^{^{''}} - 3 y^{^{'}} - 10 y = 0

\frac{1}{( 1 + x ) ^{2}}, 1

\frac{d}{d x} (5 x)

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