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

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

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

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

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

Solution steps

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

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

\frac{1}{g ^{2} ( x )}

10 x (x^{2} + 5)^{4}

y = 2 e^{3} t

\int (4 x (3 x - 2)) d x

\frac{d}{d x} (2 a arccos (2 x))

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