If the solution curve f(x,y)=0 of the differential equation (1+ _ex)(dx)/(dy)-x _ex=e^y,x…

Question

If the solution curve $f(x,y)=0$ of the differential equation

$(1+{\log }_{e}x)\frac{dx}{dy}-x{\log }_{e}x={e}^{y},x>0$,

passes through the points $(1,0)$ and $(\alpha ,2)$, then ${\alpha }^{\alpha }$ is equal to :

Accepted Answer

We have, $(1+{\log }_{e}x)\frac{dx}{dy}-x{\log }_{e}x={e}^{y},x>0$

Put $x{\log }_{e}x=t$

$\begin{gathered} \Rightarrow (x\cdot \frac{1}{x}+{\log }_{e}x)\frac{dx}{dy}=\frac{dt}{dy} \ \Rightarrow (1+{\log }_{e}x)\frac{dx}{dy}=\frac{dt}{dy} \ 	herefore \frac{dt}{dy}-t={e}^{y} \end{gathered}$

Now, IF $={e}^{\int -dy}={e}^{-y}$

$\begin{gathered} 	herefore t({e}^{-y})=\int ({e}^{y}\cdot {e}^{-y})dy+c \ \Rightarrow t{e}^{-y}=\int dy+c \ \Rightarrow t{e}^{-y}=y+c \ \Rightarrow (x{\log }_{e}x){e}^{-y}=y+c..........(i) \end{gathered}$

Equation (i), passes through the point $(1,0)$

$\begin{gathered} 	herefore 0=0+c \ \Rightarrow c=0 \ 	herefore (x{\log }_{e}x){e}^{-y}=y \ \Rightarrow x{\log }_{e}x=y{e}^{y} \end{gathered}$

Which passes through $(\alpha ,2)$

$\begin{gathered} 	herefore \alpha {\log }_{e}\alpha =2{e}^{2} \ \Rightarrow {\log }_{e}{\alpha }^{\alpha }=2{e}^{2} \ \Rightarrow {\alpha }^{\alpha }={e}^{2{e}^{2}} \end{gathered}$

Solutions · 1