Differential Equations and Linear Algebra (4th Edition)

$y=C_1(x^2+3)+1$
Multiply the entire equation by $\frac{dx}{y-1}$ to separate variables. $$\frac{dx}{y-1}(\frac{dy}{dx}=\frac{2x(y-1)}{x^2+3})$$ $$\frac{dy}{y-1}=\frac{2x}{x^2+3} dx$$ Since each side of the equation is in terms of a variable, you can integrate. $$\int \frac{dy}{y-1}=\int \frac{2x}{x^2+3}dx$$ Use a u-sub of $u=x^2+3$ and $du=2xdx$ to solve the second integral. $$ln|y-1|=ln|x^2+3|+C$$ Solve for $y$. $$y-1=e^{ln(x^2+3)+C}$$ $$y-1=e^{C}(x^2+3)$$ $$y-1=C_1(x^2+3)$$ $$y=C_1(x^2+3)+1$$