Verified Solution • 2025–26 Session moderate

QJEEM23Ma8766: JEE Main 2023 Mathematics - Class 12 JEEM23Ma8766 Solution: Let be a solution curve of the differential equation. . If the…

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

(1−x2y2)dx=ydx+xdy\left(1-x^{2} y^{2}\right) d x=y d x+x d y .

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

Multiple Choice Options

  • A 1+3e22(3e2−1)\frac{1+3 e^{2}}{2\left(3 e^{2}-1\right)}
  • B 3e22(3e2−1)\frac{3 e^{2}}{2\left(3 e^{2}-1\right)}
  • C 1−3e22(3e2+1)\frac{1-3 e^{2}}{2\left(3 e^{2}+1\right)}
  • D 3e22(3e2+1)\frac{3 e^{2}}{2\left(3 e^{2}+1\right)}

Step-by-Step Verified Solution

We have,

amp;(1−x2y2)dx=ydx+xdy,y(1)=2amp;dx=ydx+xdy1−(xy)2\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

∫dxamp;=∫d(xy)1−(xy)2xamp;=12log⁡∣1+xy1−xy∣+C\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)=2y(1)=2

1amp;=12log⁡∣1+21−2∣+C⇒Camp;=1−12log⁡3\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=2x=2 as y(2)=αy(2)=\alpha

2=12log⁡∣1+2α1−2α∣+1−12log⁡32=\frac{1}{2} \log \left|\frac{1+2 \alpha}{1-2 \alpha}\right|+1-\frac{1}{2} \log 3

1+12log⁡3amp;=12log⁡∣1+2α1−2α∣2+log⁡3amp;=log⁡∣1+2α1−2α∣\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}

amp;⇒1+2α1−2α∣=3e2amp;⇒1+2α1−2α=±3e21amp;⇒12α=±3e2+1±3e2−1amp;⇒2α=±3e2−1±3e2+1\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}

⇒α=12(±3e2−11±3e2)∴α=3e2−12(1+3e2) or 3e2+12(3e2−1)\begin{array}{ll}\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{array}

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