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 $\frac{d y}{d x}=0$

$\Rightarrow x=0, \pm 1$

JEE Advanced 2023 Paper 2 Online Mathematics - Differential Equations Question 6 English Explanation

JEE Advanced 2023 Paper 2 Online Mathematics - Differential Equations Question 6 English Explanation

From (1),

$$ y_{\text {max }}=16 $$19:T60c,$\begin{aligned} &\left(x^2-5\right) \frac{d y}{d x}-2 x y =-2 x\left(x^2-5\right)^2 \\ \\&\Rightarrow \frac{d y}{d x}-\frac{2 x}{\left(x^2-5\right)} y=-2 x\left(x^2-5\right)\end{aligned}$

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

अब

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

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

अतः,

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

Differential Equations

Continue this paper

Similar Differential Equations PYQs

More practice from this question