What is the solution for (dy)/(dx)=(2x(y-1))/(x^2+3),y(1)=9 ?

The solution for (dy)/(dx)=(2x(y-1))/(x^2+3),y(1)=9 is y=2x^2+7

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(dy)/(dx)=(2x(y-1))/(x^2+3),y(1)=9

\frac{d y}{d x} = \frac{2 x ( y - 1 )}{x ^{2} + 3}, y (1) = 9

y = 2 x^{2} + 7

Solution steps

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

Solve separable ODE:

y = 2 x^{2} + 7

y = 2 x^{2} + 7

\int \frac{x}{( x ^{2} + 1 ) ^{4}} d x

2 t \cdot cos (t)

\int 3 t cos (2 t) d t

x \to 8 lim (\frac{∣ 8 - x ∣}{8 - x})

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

What is the solution for (dy)/(dx)=(2x(y-1))/(x^2+3),y(1)=9 ?
The solution for (dy)/(dx)=(2x(y-1))/(x^2+3),y(1)=9 is y=2x^2+7