## Calculus (3rd Edition)

$$y=-\sqrt{\ln(x^2+e^4)}.$$
By separation of variables, we have $$ye^{y^2}dy=xdx$$ then by integration, we get $$\frac{1}{2}e^{y^2}=\frac{1}{2}x^2+c\Longrightarrow y^2=\ln(x^2+A).$$ Now, since $y(0)=-2$, then $4=\ln(0+A)\Longrightarrow A= e^{4}$. So the general solution is given by $$y^2=\ln(x^2+e^4).$$ We take the square root to get: $$y=-\sqrt{\ln(x^2+e^4)}.$$ Note that we use the negative value because we know that $y(0)$ is negative.