QJEEAdv23Ma9c9c: JEE Advanced 2023 Mathematics - Class 12 JEEAdv23Ma9c9c Solution: For , let be a solution of the differential equation such that…

Step-by-Step Verified Solution

$\begin{aligned} &\left(x^2-5\right) \frac{d y}{d x}-2 x y =-2 x\left(x^2-5\right)^2 \\ \\& \Rightarrow \frac{d y}{d x}-\frac{2 x}{\left(x^2-5\right)} y=-2 x\left(x^2-5\right)\end{aligned}$

I.F. $=e^{-\int \frac{2 x}{x^2-5} d x}=\frac{1}{\left(x^2-5\right)}$

Now

$\begin{aligned}y \cdot \frac{1}{\left(x^2-5\right)} & =-\int 2 x d x \\\\& =-x^2+c\end{aligned}$

$\begin{array}{rlrl} &\Rightarrow y =c\left(x^2-5\right)-x^2\left(x^2-5\right) \\\\ & y(2) =7 \\\\ &\Rightarrow 7 =-c+4 \\\\ &\Rightarrow c=-3\end{array}$

So,

$\begin{aligned}y & =\left(x^2-5\right)\left(-x^2-3\right) ............(1) \\\\\frac{d y}{d x} & =\left(x^2-5\right)(-2 x)+\left(-x^2-3\right)(2 x) \\\\& =2 x\left(-x+5-x^2-3\right) \\\\& =2 x\left(-2 x^2+2\right)\end{aligned}$

For maxima and minima, put $\frac{d y}{d x}=0$

$\Rightarrow x=0, \pm 1$


From (1),

$y_{\text {max }}=16$