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Differential Equations question

2025 · Shift 1 · Q29
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  5. /2025 · Shift 1 · Q29

Differential Equations question

2025 · Shift 1 · Q29

JEE AdvancedMathematicsDifferential EquationsNumerical+4 / −1
For all x > 0, let y₁(x), y₂(x), and y₃(x) be the functions satisfying dy1dx−(sin⁡x)2y1=0,y1(1)=5,dy2dx−(cos⁡x)2y2=0,y2(1)=13,dy3dx−(2−x3)x3y3=0,y3(1)=35e,\frac{dy_1}{dx} - (\sin x)^2 y_1 = 0, \quad y_1(1) = 5, \frac{dy_2}{dx} - (\cos x)^2 y_2 = 0, \quad y_2(1) = \frac{1}{3}, \frac{dy_3}{dx} - \frac{(2-x^3)}{x^3} y_3 = 0, \quad y_3(1) = \frac{3}{5e},dxdy1​​−(sinx)2y1​=0,y1​(1)=5,dxdy2​​−(cosx)2y2​=0,y2​(1)=31​,dxdy3​​−x3(2−x3)​y3​=0,y3​(1)=5e3​, respectively. Then lim⁡x→0+y1(x)y2(x)y3(x)+2xe3xsin⁡x\lim\limits_{x \to 0^+} \frac{y_1(x)y_2(x)y_3(x) + 2x}{e^{3x} \sin x}x→0+lim​e3xsinxy1​(x)y2​(x)y3​(x)+2x​ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. Write each differential equation in separable/standard linear form

We have dy1dx=(sin⁡x)2y1,dy2dx=(cos⁡x)2y2,dy3dx=2−x3x3y3.\frac{dy_1}{dx}=(\sin x)^2y_1,\qquad \frac{dy_2}{dx}=(\cos x)^2y_2,\qquad \frac{dy_3}{dx}=\frac{2-x^3}{x^3}y_3.dxdy1​​=(sinx)2y1​,dxdy2​​=(cosx)2y2​,dxdy3​​=x32−x3​y3​.

So each satisfies yi′yi=fi(x),\frac{y_i'}{y_i}=f_i(x),yi​yi′​​=fi​(x), which gives yi(x)=C e∫fi(x) dx.y_i(x)=C\,e^{\int f_i(x)\,dx}.yi​(x)=Ce∫fi​(x)dx.

  1. Find the product y1y2y3y_1y_2y_3y1​y2​y3​ directly

Instead of solving each completely, observe:

ddxln⁡(y1y2y3)=y1′y1+y2′y2+y3′y3.\frac{d}{dx}\ln(y_1y_2y_3)=\frac{y_1'}{y_1}+\frac{y_2'}{y_2}+\frac{y_3'}{y_3}.dxd​ln(y1​y2​y3​)=y1​y1′​​+y2​y2′​​+y3​y3′​​.

Hence ddxln⁡(y1y2y3)=(sin⁡x)2+(cos⁡x)2+2−x3x3.\frac{d}{dx}\ln(y_1y_2y_3)=(\sin x)^2+(\cos x)^2+\frac{2-x^3}{x^3}.dxd​ln(y1​y2​y3​)=(sinx)2+(cosx)2+x32−x3​.

Now (sin⁡x)2+(cos⁡x)2=1,(\sin x)^2+(\cos x)^2=1,(sinx)2+(cosx)2=1, and 2−x3x3=2x3−1.\frac{2-x^3}{x^3}=\frac{2}{x^3}-1.x32−x3​=x32​−1.

Therefore, ddxln⁡(y1y2y3)=1+(2x3−1)=2x3.\frac{d}{dx}\ln(y_1y_2y_3)=1+\left(\frac{2}{x^3}-1\right)=\frac{2}{x^3}.dxd​ln(y1​y2​y3​)=1+(x32​−1)=x32​.

So ln⁡(y1y2y3)=∫2x3 dx=−1x2+C.\ln(y_1y_2y_3)=\int \frac{2}{x^3}\,dx=-\frac{1}{x^2}+C.ln(y1​y2​y3​)=∫x32​dx=−x21​+C.

Thus y1y2y3=Ae−1/x2y_1y_2y_3=Ae^{-1/x^2}y1​y2​y3​=Ae−1/x2 for some constant AAA.

  1. Use the initial conditions to determine AAA

At x=1x=1x=1, y1(1)y2(1)y3(1)=5⋅13⋅35e=1e.y_1(1)y_2(1)y_3(1)=5\cdot \frac13\cdot \frac{3}{5e}=\frac{1}{e}.y1​(1)y2​(1)y3​(1)=5⋅31​⋅5e3​=e1​.

But from the product form, y1(1)y2(1)y3(1)=Ae−1.y_1(1)y_2(1)y_3(1)=Ae^{-1}.y1​(1)y2​(1)y3​(1)=Ae−1.

So Ae−1=1e  ⟹  A=1.Ae^{-1}=\frac1e \implies A=1.Ae−1=e1​⟹A=1.

Hence y1(x)y2(x)y3(x)=e−1/x2.y_1(x)y_2(x)y_3(x)=e^{-1/x^2}.y1​(x)y2​(x)y3​(x)=e−1/x2.

  1. Evaluate the limit

We need

=\lim_{x\to 0^+}\frac{e^{-1/x^2}+2x}{e^{3x}\sin x}.$$ As $x\to 0^+$, $$e^{-1/x^2}\to 0$$ very rapidly. In particular, $$\frac{e^{-1/x^2}}{x}\to 0.$$ Also, $$e^{3x}\to 1, \qquad \frac{\sin x}{x}\to 1.$$ So $$e^{3x}\sin x \sim x.$$ Therefore, $$L=\lim_{x\to 0^+}\left(\frac{e^{-1/x^2}}{e^{3x}\sin x}+\frac{2x}{e^{3x}\sin x}\right).$$ Now, $$\frac{e^{-1/x^2}}{e^{3x}\sin x}\to 0,$$ and $$\frac{2x}{e^{3x}\sin x}=2\cdot \frac{x}{\sin x}\cdot e^{-3x}\to 2\cdot 1\cdot 1=2.$$ Thus, $$L=2.$$ --- 5. **Final answer** The required integer is $$\boxed{2}.$$ This matches the stored correct answer.
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