If y=y(x) solves (1+e^2x)dydx+2(1+y^2)e^x=0 and y(0)=0, then 6(y'(0)+(y(_e√3))^2) is:

Question

If $y=y(x)$ solves $(1+e^{2x})\dfrac{dy}{dx}+2(1+y^2)e^x=0$ and $y(0)=0$, then $6\left(y'(0)+(y(\log_e\sqrt3))^2\right)$ is:

Accepted Answer

$$\frac{dy}{1+y^2}+\frac{2e^x}{1+e^{2x}}\,dx = 0.$$

Integrating: $\tan^{-1}y+2\tan^{-1}(e^x)=c$. Using $y(0)=0$: $c=\pi/2$.

From the original ODE at $x=0$: $y'(0)=-1$. At $x=\ln\sqrt3$: $y(\ln\sqrt3)=-1/\sqrt3$.

$$6\left(-1+\frac13\right) = -4.$$

Solutions · 1