For all \(x>0\), let \(y_{1}(x), y_{2}(x)\), and \(y_{3}(x)\) be the functions satisfying \[ \begin{gathered} \frac{d y_{1}}{d x}-(\sin x)^{2} y_{1}=0, y_{1}(1)=5 \\ \frac{d y_{2}}{d x}-(\cos x)^{2} y_{2}=0, y_{2}(1)=\frac{1}{3} \\ \frac{d y_{3}}{d x}-\left(\frac{2-x^{3}}{x^{3}}\right) y_{3}=0, y_{3}(1)=\frac{3}{5 e} \end{gathered} \] respectively. Then \[ \lim _{x \rightarrow 0^{+}} \frac{y_{1}(x) y_{2}(x) y_{3}(x)+2 x}{e^{3 x} \sin x} \] is equal to ____.

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Differential Equations

For all

x > 0

, let

y_{1} (x), y_{2} (x)

, and

y_{3} (x)

be the functions satisfying

\frac{d y _{1}}{d x} - (sin x)^{2} y_{1} = 0, y_{1} (1) = 5 \frac{d y _{2}}{d x} - (cos x)^{2} y_{2} = 0, y_{2} (1) = \frac{1}{3} \frac{d y _{3}}{d x} - (\frac{2 - x ^{3}}{x ^{3}}) y_{3} = 0, y_{3} (1) = \frac{3}{5 e}

respectively. Then

x \to 0^{+} lim \frac{y _{1} ( x ) y _{2} ( x ) y _{3} ( x ) + 2 x}{e ^{3 x} sin x}

is equal to ____.

Answer: 2

Adding the three differential equations:

\frac{d y _{1}}{y _{1}} + \frac{d y _{2}}{y _{2}} + \frac{d y _{3}}{y _{3}} = (sin^{2} x + cos^{2} x + \frac{2 - x ^{3}}{x ^{3}}) d x

= (1 + \frac{2}{x ^{3}} - 1) d x = \frac{2}{x ^{3}} d x

Integrating:

ln (y_{1} y_{2} y_{3}) = - \frac{1}{x ^{2}} + C

Using initial conditions at

x = 1

ln (5 \cdot \frac{1}{3} \cdot \frac{3}{5 e}) = - 1 + C

ln (\frac{1}{e}) = - 1 + C

- 1 = - 1 + C ⟹ C = 0

Therefore:

y_{1} (x) y_{2} (x) y_{3} (x) = e^{- 1/ x^{2}}

The limit becomes:

x \to 0^{+} lim \frac{e ^{- 1/ x^{2}} + 2 x}{e ^{3 x} sin x}

Split into two parts:

x \to 0^{+} lim \frac{e ^{- 1/ x^{2}}}{e ^{3 x} sin x} + x \to 0^{+} lim \frac{2 x}{e ^{3 x} sin x}

First limit: As

x \to 0^{+}

e^{- 1/ x^{2}} \to 0

much faster than any polynomial or exponential, so this limit is 0.

Second limit:

x \to 0^{+} lim \frac{2 x}{e ^{3 x} sin x} = x \to 0^{+} lim \frac{2 x}{( 1 + 3 x + \dots ) ( x - \dots )} = x \to 0^{+} lim \frac{2 x}{x} = 2

Total:

0 + 2 = 2

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer