Let y=y(x) be a solution curve of the differential equation.

Question

Let $y=y(x)$ be a solution curve of the differential equation.

$(1-{x}^{2}{y}^{2})dx=ydx+xdy$.

If the line $x=1$ intersects the curve $y=y(x)$ at $y=2$ and the line $x=2$ intersects the curve $y=y(x)$ at $y=\alpha $, then a value of $\alpha $ is :

Accepted Answer

We have,

$\begin{gathered} (1-{x}^{2}{y}^{2})dx=ydx+xdy,y(1)=2 \ dx=\frac{ydx+xdy}{1-(xy{)}^{2}} \end{gathered}$

On integrating both sides, we get

$\begin{gathered} \int dx =\int \frac{d(xy)}{1-(xy{)}^{2}} \ \ x =\frac{1}{2}\log |\frac{1+xy}{1-xy}|+C $

As, $y(1)=2$

$ 1 =\frac{1}{2}\log |\frac{1+2}{1-2}|+C \ \ \Rightarrow C =1-\frac{1}{2}\log 3 $

Now, substitute $x=2$ as $y(2)=\alpha $

$2=\frac{1}{2}\log |\frac{1+2\alpha }{1-2\alpha }|+1-\frac{1}{2}\log 3$

$ 1+\frac{1}{2}\log 3 =\frac{1}{2}\log |\frac{1+2\alpha }{1-2\alpha }| \ \ 2+\log 3 =\log |\frac{1+2\alpha }{1-2\alpha }| \end{gathered}$

$\begin{gathered} \Rightarrow \frac{1+2\alpha }{1-2\alpha }|=3{e}^{2} \ \Rightarrow \frac{1+2\alpha }{1-2\alpha }=\pm \frac{3{e}^{2}}{1} \ \Rightarrow \frac{1}{2\alpha }=\frac{\pm 3{e}^{2}+1}{\pm 3{e}^{2}-1} \ \Rightarrow 2\alpha =\frac{\pm 3{e}^{2}-1}{\pm 3{e}^{2}+1} \end{gathered}$

$\begin{gathered} \Rightarrow \alpha =\frac{1}{2}(\frac{\pm 3{e}^{2}-1}{1\pm 3{e}^{2}}) \ 	herefore \alpha =\frac{3{e}^{2}-1}{2(1+3{e}^{2})}\frac{3{e}^{2}+1}{2(3{e}^{2}-1)} \end{gathered}$

Solutions · 1