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  • Let y = y(x) be a solution curve of the differential equation.<img alt="\left(1-x^2 y^2\right) d x=y d x+x d y" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%281-x%5E2%20y%5E2%5Crigh

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

Option: 1

\frac{1+3 e^2}{2\left(3 e^2-1\right)}


Option: 2

\frac{1-3 \mathrm{e}^2}{2\left(3 \mathrm{e}^2+1\right)}


Option: 3

\frac{3 \mathrm{e}^2}{2\left(3 \mathrm{e}^2-1\right)}


Option: 4

\frac{3 \mathrm{e}^2}{2\left(3 \mathrm{e}^2+1\right)}


Answers (1)

best_answer

\begin{aligned} & \left(1-x^2 y^2\right) d x=y d x+x d y, y(1)=2 \\ & y(2)=\infty \\ & d x=\frac{d(x y)}{1-(x y)^2} \\ & \int d x=\int \frac{d(x y)}{1-(x y)^2} \\ & x=\frac{1}{2} \ln \left|\frac{1+x y}{1-x y}\right|+C \\ & \text { Put } x=1 \text { and } y=2 \text { : } \\ & 1=\frac{1}{2} \ln \left|\frac{1+2}{1-2}\right|+C \\ & C=1-\frac{1}{2} \ln 3 \end{aligned}

Now put $x=2$ : $$ \begin{aligned} & 2=\frac{1}{2} \ln \left|\frac{1+2 \alpha}{1-2 \alpha}\right|+1-\frac{1}{2} \ln 3 \\ & 1+\frac{1}{2} \ln 3=\frac{1}{2}\left|\frac{1+2 \alpha}{1-2 \alpha}\right| \\ & 2+\ln 3=\left|\frac{1+2 \alpha}{1-2 \alpha}\right| \\ & \left|\frac{1+2 \alpha}{1-2 \alpha}\right|=3 \mathrm{e}^2 \\ & \frac{1+2 \alpha}{1-2 \alpha}=3 \mathrm{e}^2,-3 \mathrm{e}^2 \\ & \frac{1+2 \alpha}{1-2 \alpha}=3 \mathrm{e}^2 \Rightarrow \alpha=\frac{3 \mathrm{e}^2-1}{2\left(3 \mathrm{e}^2+1\right)} \\ & \text { And } \frac{1+2 \alpha}{1-2 \alpha}=-3 \mathrm{e}^2 \Rightarrow \alpha=\frac{3 \mathrm{e}^2+1}{2\left(3 \mathrm{e}^2-1\right)} \end{aligned}

Posted by

Irshad Anwar

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