JEE Main 2024 — 30 January, Shift 1 — Question 27

Let the given differential equation be

$$(1 - x^2)dy = [xy + (x^3 + 2)\sqrt{3(1 - x^2)}]dx$$

The domain for $x$ is $-1 < x < 1$. Rearrange the equation into the standard linear first-order differential equation form, $\frac{dy}{dx} + P(x)y = Q(x)$: Divide by $dx$: $(1 - x^2)\frac{dy}{dx} = xy + (x^3 + 2)\sqrt{3(1 - x^2)}$ Divide by $(1 - x^2)$: $\frac{dy}{dx} = \frac{xy}{1 - x^2} + \frac{(x^3 + 2)\sqrt{3(1 - x^2)}}{1 - x^2}$ $\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{(x^3 + 2)\sqrt{3}}{\sqrt{1 - x^2}}$ Here, $P(x) = -\frac{x}{1 - x^2}$ and $Q(x) = \frac{(x^3 + 2)\sqrt{3}}{\sqrt{1 - x^2}}$.

Calculate the integrating factor (IF), which is $e^{\int P(x)dx}$: $\int P(x)dx = \int -\frac{x}{1 - x^2}dx$ Let $u = 1 - x^2$, so $du = -2xdx \implies xdx = -\frac{1}{2}du$. $\int -\frac{x}{1 - x^2}dx = \int \frac{1}{u} \cdot \frac{1}{2}du = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(1 - x^2)$ (since $1-x^2 > 0$ for $-1 < x < 1$) $\int P(x)dx = \ln((1 - x^2)^{1/2}) = \ln(\sqrt{1 - x^2})$. So, the integrating factor is $\text{IF} = e^{\ln(\sqrt{1 - x^2})} = \sqrt{1 - x^2}$.

Multiply the differential equation by the integrating factor: $\sqrt{1 - x^2}\left(\frac{dy}{dx} - \frac{x}{1 - x^2}y\right) = \sqrt{1 - x^2}\left(\frac{(x^3 + 2)\sqrt{3}}{\sqrt{1 - x^2}}\right)$ $\frac{d}{dx}(y\sqrt{1 - x^2}) = (x^3 + 2)\sqrt{3}$

Integrate both sides with respect to $x$: $\int \frac{d}{dx}(y\sqrt{1 - x^2})dx = \int (x^3 + 2)\sqrt{3}dx$ $y\sqrt{1 - x^2} = \sqrt{3} \left( \frac{x^4}{4} + 2x \right) + C$

Use the initial condition $y(0) = 0$ to find the constant $C$: $0 \cdot \sqrt{1 - 0^2} = \sqrt{3} \left( \frac{0^4}{4} + 2(0) \right) + C$ $0 = 0 + C \implies C = 0$.

The particular solution is: $y\sqrt{1 - x^2} = \sqrt{3} \left( \frac{x^4}{4} + 2x \right)$ $y(x) = \frac{\sqrt{3} \left( \frac{x^4}{4} + 2x \right)}{\sqrt{1 - x^2}}$

We need to find $y(\frac{1}{2}) = \frac{m}{n}$: Substitute $x = \frac{1}{2}$: $y\left(\frac{1}{2}\right) = \frac{\sqrt{3} \left( \frac{(1/2)^4}{4} + 2(1/2) \right)}{\sqrt{1 - (1/2)^2}}$ $y\left(\frac{1}{2}\right) = \frac{\sqrt{3} \left( \frac{1/16}{4} + 1 \right)}{\sqrt{1 - 1/4}}$ $y\left(\frac{1}{2}\right) = \frac{\sqrt{3} \left( \frac{1}{64} + 1 \right)}{\sqrt{3/4}}$ $y\left(\frac{1}{2}\right) = \frac{\sqrt{3} \left( \frac{1+64}{64} \right)}{\frac{\sqrt{3}}{2}}$ $y\left(\frac{1}{2}\right) = \frac{\sqrt{3} \left( \frac{65}{64} \right)}{\frac{\sqrt{3}}{2}}$ $y\left(\frac{1}{2}\right) = \frac{65}{64} \cdot \frac{2}{1} = \frac{65}{32}$

So, $y(\frac{1}{2}) = \frac{65}{32}$. This means $m=65$ and $n=32$. We need to verify if $m$ and $n$ are coprime numbers. Prime factorization of $65 = 5 \times 13$. Prime factorization of $32 = 2^5$. Since there are no common prime factors, 65 and 32 are coprime.

Finally, calculate $m+n$: $m+n = 65 + 32 = 97$.

The final answer is $\boxed{97}$.