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.

$$\left(1-x^{2} y^{2}\right) d x=y d x+x d y$$.

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,

$$
\begin{aligned}
& \left(1-x^2 y^2\right) d x=y d x+x d y, y(1)=2 \\
& d x=\frac{y d x+x d y}{1-(x y)^2}
\end{aligned}
$$

On integrating both sides, we get

$$
\begin{aligned}
\int d x & =\int \frac{d(x y)}{1-(x y)^2} \\
x & =\frac{1}{2} \log \left|\frac{1+x y}{1-x y}\right|+C
\end{aligned}
$$

As, $y(1)=2$

$$
\begin{aligned}
1 & =\frac{1}{2} \log \left|\frac{1+2}{1-2}\right|+C \\
\Rightarrow C & =1-\frac{1}{2} \log 3
\end{aligned}
$$

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

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

$$
\begin{aligned}
1+\frac{1}{2} \log 3 & =\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right| \\
2+\log 3 & =\log \left|\frac{1+2 \alpha}{1-2 \alpha}\right|
\end{aligned}
$$

$$
\begin{aligned}
&\Rightarrow \frac{1+2 \alpha}{1-2 \alpha} \mid  =3 e^2 \\
&\Rightarrow \frac{1+2 \alpha}{1-2 \alpha}= \pm \frac{3 e^2}{1} \\
& \Rightarrow\frac{1}{2 \alpha}  =\frac{ \pm 3 e^2+1}{ \pm 3 e^2-1} \\
&\Rightarrow 2 \alpha=\frac{ \pm 3 e^2-1}{ \pm 3 e^2+1}
\end{aligned}
$$

$$
\[\begin{gathered} \Rightarrow  \alpha=\frac{1}{2}\left(\frac{ \pm 3 e^2-1}{1 \pm 3 e^2}\right) \ \therefore  \alpha=\frac{3 e^2-1}{2\left(1+3 e^2\right)} \text { or } \frac{3 e^2+1}{2\left(3 e^2-1\right)} \end{gathered}$$
\]

Solutions · 1