If the solution curve of the differential equation (y-2 _ex)dx+(x _ex^2)dy=0,x1 passes th…

Question

If the solution curve of the differential equation $(y-2{\log }_{e}x)dx+(x{\log }_{e}{x}^{2})dy=0,x>1$ passes through the points $(e,\frac{4}{3})$ and $({e}^{4},\alpha )$, then $\alpha $ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

The given differential equation is,

$\begin{gathered} (y-2\log x)dx+(x\log {x}^{2})dy=0 \ \Rightarrow \frac{dy}{dx}=\frac{(2\log x-y)}{2x\log x} \ \Rightarrow \frac{dy}{dx}+\frac{y}{2x\log x}=\frac{1}{x} \end{gathered}$

It is a linear differential equation.

$	herefore ={e}^{\int \frac{1}{2x\log x}dx}$

Put $\log x=t\Rightarrow \frac{1}{x}dx=dt$

$	herefore ={e}^{\int \frac{1}{2t}dt}={e}^{\log (t{)}^{\frac{1}{2}}}=\sqrt{t}=\sqrt{\log x}$

So, required solution is,

$\begin{gathered} y\sqrt{\log x}=\int \frac{\sqrt{\log x}}{x}dx \ \log x=v\Rightarrow \frac{1}{x}dx=dv \ \Rightarrow y\sqrt{\log x}=\int \sqrt{v}dv+C \ \Rightarrow y\sqrt{\log x}=\frac{2{v}^{3/2}}{3}+C \ \Rightarrow y\sqrt{\log x}=\frac{2}{3}(\log x{)}^{3/2}+C \end{gathered}$

Now, this curve passes through $(e,\frac{4}{3})$ and $({e}^{4},\alpha )$

$\begin{gathered} 	herefore \frac{4}{3}\sqrt{\log e}=\frac{2}{3}(\log e{)}^{3/2}+C \ \Rightarrow C=\frac{4}{3}-\frac{2}{3}=\frac{2}{3} \end{gathered}$

Also, $\alpha \sqrt{\log {e}^{4}}=\frac{2}{3}{(\log {e}^{4})}^{3/2}+\frac{2}{3}$

$\begin{gathered} \Rightarrow 2\alpha =\frac{2}{3}	imes (4{)}^{3/2}+\frac{2}{3}=\frac{16}{3}+\frac{2}{3}=\frac{18}{3} \ \Rightarrow \alpha =3 \end{gathered}$

Solutions · 1