JEE Advanced 2025 Paper 1 · Q13 · Differential Equations

Question

For $x > 0$, let $y_1, y_2, y_3$ satisfy $\frac{dy_1}{dx} - (\sin x)^2 y_1 = 0$ with $y_1(1) = 5$; $\frac{dy_2}{dx} - (\cos x)^2 y_2 = 0$ with $y_2(1) = \frac{1}{3}$; $\frac{dy_3}{dx} - \frac{2 - x^3}{x^3} y_3 = 0$ with $y_3(1) = \frac{3}{5e}$. Find $\displaystyle \lim_{x 	o 0^{+}} \frac{y_1(x) y_2(x) y_3(x) + 2x}{e^{3x} \sin x}$.

SolveKar AI · Accepted Answer

Each ODE is $\frac{dy}{dx} = P(x) y$, so $y(x) = y(x_0) \exp\left(\int_{x_0}^{x} P(s)\, dsight)$. Multiplying $y_1 \cdot y_2 \cdot y_3$ gives an exponential of the sum of the three integrands integrated from $1$ to $x$. Sum: $\sin^2(s) + \cos^2(s) + \frac{2 - s^3}{s^3} = 1 + \frac{2}{s^3} - 1 = \frac{2}{s^3}$. Integral of $\frac{2}{s^3}$ from $1$ to $x = -\frac{1}{x^2} + 1$. So $y_1 y_2 y_3 = \left(5 \cdot \frac{1}{3} \cdot \frac{3}{5e}ight) \cdot \exp\left(1 - \frac{1}{x^2}ight) = \frac{1}{e} \cdot \exp\left(1 - \frac{1}{x^2}ight) = \exp\left(-\frac{1}{x^2}ight)$. As $x 	o 0^{+}$, $\exp\left(-\frac{1}{x^2}ight) 	o 0$ faster than any polynomial. Numerator: $y_1 y_2 y_3 + 2x \approx 2x$ (the exponential term is negligible). Denominator: $e^{3x} \sin x \approx (1)(x) = x$. Limit $= 2x / x = 2$. Answer: $2$.

JEE Advanced 2025 Paper 1 · Q13 · Differential Equations

Solution