JEE Main 2024 — 9 April, Shift 1 — Question 17

Given: $(x^2 + y^2) dx - 5xy\, dy = 0$, $y(1) = 0$ Rewrite as: $\frac{dy}{dx} = \frac{x^2 + y^2}{5xy}$ Substitute $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$ Then: $v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{5x(vx)} = \frac{1 + v^2}{5v}$ Rearrange: $x\frac{dv}{dx} = \frac{1 + v^2}{5v} - v = \frac{1 + v^2 - 5v^2}{5v} = \frac{1 - 4v^2}{5v}$ Separate variables: $\frac{5v}{1 - 4v^2} dv = \frac{dx}{x}$ Integrate: $\int \frac{5v}{1 - 4v^2} dv = \int \frac{dx}{x}$ Let $t = 1 - 4v^2$, then $dt = -8v\, dv$ or $v\, dv = -\frac{dt}{8}$ $\int \frac{5}{t} \cdot \left(-\frac{dt}{8}\right) = \ln|x| + C$ $-\frac{5}{8} \ln|t| = \ln|x| + C$ Multiply by 8: $-5\ln|t| = 8\ln|x| + 8C$ Exponentiate: $|t|^{-5} = K x^8$, where $K = e^{8C}$ Substitute back: $|1 - 4v^2|^{-5} = K x^8$ or $|1 - 4v^2|^{5} = \frac{1}{K} x^{-8}$ Let $C' = 1/K$: $|1 - 4v^2|^{5} = C' x^{-8}$ Replace $v = y/x$: $\left|1 - \frac{4y^2}{x^2}\right|^{5} = C' x^{-8}$ Multiply by $x^{10}$: $|x^2 - 4y^2|^{5} = C' x^{2}$ Apply $y(1) = 0$: $|1 - 0|^{5} = C' \cdot 1 \Rightarrow C' = 1$ Thus: $|x^2 - 4y^2|^{5} = x^{2}$