JEE Advanced 2025 Paper 1, Mathematics Section 3 Q6: Methods of Solving First Order, First Degree Differential Equation

Each ODE is separable: $\frac{d{y}_i}{y_i}=P_i(x)dx$. Let $Y=y_1{y}_2{y}_3$. Then

$\frac{Y^{\prime }}{Y}=\frac{y_1^{\prime }}{y_1}+\frac{y_2^{\prime }}{y_2}+\frac{y_3^{\prime }}{y_3}={\sin }^{2}x+{\cos }^{2}x+\frac{2-x^3}{x^3}=1+\frac{2}{x^3}-1=\frac{2}{x^3}.$

Integrating, $\ln Y=-\frac{1}{x^2}+C.$

Apply the initial conditions at $x=1$: $Y(1)=5\cdot \frac{1}{3}\cdot \frac{3}{5e}=\frac{1}{e}$, so $\ln (1/e)=-1=-1+C\Rightarrow C=0$.

Therefore $Y(x)=e^{-1/x^2}$.

The limit becomes $\underset{x\to 0^{+}}{\lim }\frac{e^{-1/x^2}+2x}{e^{3x}\sin x}.$

Split into two parts. The first part is $\frac{e^{-1/x^2}}{e^{3x}\sin x}.$ As $x\to 0^{+}$, $e^{-1/x^2}$ decays to $0$ faster than any polynomial, while the denominator behaves like $x$. Hence this part $\to 0$.

The second part is $\frac{2x}{e^{3x}\sin x}=\frac{2}{e^{3x}}\cdot \frac{x}{\sin x}\to \frac{2}{1}\cdot 1=2.$

Total limit $=0+2=\mathrm{2.00.}$

$\therefore$ The answer is $\boxed{2.00}$.