JEE Main 2023 — Differential Equations Question with Solution

We can solve the given differential equation using an integrating factor.

The integrating factor is given by :

$$\mu (x)=e^{\int \frac{4x}{x^2-1}dx}=e^{2\ln (x^2-1)}=(x^2-1{)}^2$$

Multiplying both sides of the differential equation by $\mu (x)$, we get :

$$(x^2-1{)}^2\frac{dy}{dx}+4xy=\frac{x+2}{{\left(x^2-1\right)}^{\frac{1}{2}}}$$

We can rewrite the left-hand side using the product rule:

$$\frac{d}{dx}\left((x^2-1{)}^{2}y\right)=\frac{x+2}{{\left(x^2-1\right)}^{\frac{1}{2}}}$$

Integrating both sides with respect to $x$, we get:

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

where $C$ is the constant of integration. Using the initial condition $y(2)=\frac{2}{9}{\log }_e(2+\sqrt{3})$, we can solve for $C$:

At $x=2$,

$9\cdot \frac{2}{9}\ln (2+\sqrt{3})=2\ln (2+\sqrt{3})+\sqrt{3}+C$

$C=-\sqrt{3}$

At $x=\sqrt{2}$

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

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

$$\Rightarrow \alpha \beta \gamma =6$$