JEE Main 2023 — Differential Equations Question with Solution

Given differential equation is $\frac{dy}{dx}+\frac{4x}{\left(x^2-1\right)}y=\frac{x+2}{{\left(x^2-1\right)}^{\frac{5}{2}}},x>1$

Now $\mathrm{IF}=e^{\int \frac{4x}{x^2-1}dx}={\left(x^2-1\right)}^2$

The required equation will be $\Rightarrow y·{\left(x^2-1\right)}^2=\int \frac{x+2}{{\left(x^2-1\right)}^{1/2}}dx$

$\Rightarrow y·{\left(x^2-1\right)}^2=\frac{1}{2}\int \frac{2x}{{\left(x^2-1\right)}^{1/2}}dx+2\int \frac{dx}{{\left(x^2-1\right)}^{1/2}}$

$\Rightarrow y·{\left(x^2-1\right)}^2=2\ln \left(\left|\sqrt{x^2-1}+x\right|\right)+\sqrt{x^2-1}+C$

Now using, at $x=2,$ $y\left(2\right)=\frac{2}{9}{\log }_e\left(2+\sqrt{3}\right)$ we get, 

$\Rightarrow 9·\frac{2}{9}\ln \left(2+\sqrt{3}\right)=2\ln \left(2+\sqrt{3}\right)+\sqrt{3}+C$

$\Rightarrow C=-\sqrt{3}$

Now finding the value of function at $x=\sqrt{2}$ we get,

$y\times 1=2\ln \left(1+\sqrt{2}\right)+1-\sqrt{3}$

Now on comparing we get,

$\Rightarrow \beta =1, \alpha =2, \gamma =3$

$\Rightarrow \alpha \beta \gamma =1\times 2\times 3=6$

Hence, this is the required answer.