If the solution curve f(x,y)=0 of (1+_ex)dxdy-x_ex=e^y, x0, passes through (1,0) and (,2)…

Question

If the solution curve $f(x,y)=0$ of $(1+\log_ex)\dfrac{dx}{dy}-x\log_ex=e^y$, $x>0$, passes through $(1,0)$ and $(\alpha,2)$, then $\alpha^\alpha$ is:

Accepted Answer

Let $t=x\ln x$. Then $\dfrac{dt}{dy}-t=e^y$; I.F.$=e^{-y}$, giving $te^{-y}=y+c$.

Using $(1,0)$: $c=0$. Using $(\alpha,2)$: $(\alpha\ln\alpha)e^{-2}=2 \implies \alpha\ln\alpha=2e^2$, and working through gives $\alpha^\alpha = e^{2e^2}$.

Solutions · 1