Differential EquationshardPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Linear ODE with Integrating Factor: α = 13/12 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let y=y(x)y=y(x) be the solution of dydx+(6x2+(3x2+2x3+4)e−2x(x3+2)(2+e−2x))y=2+e−2x\dfrac{dy}{dx}+\left(\dfrac{6x^2+(3x^2+2x^3+4)e^{-2x}}{(x^3+2)(2+e^{-2x})}\right)y=2+e^{-2x}, x∈(−1,2)x\in(-1,2), satisfying y(0)=32y(0)=\dfrac32. If y(1)=α(2+e−2)y(1)=\alpha(2+e^{-2}), then α\alpha is equal to
A138\dfrac{13}{8}
B613\dfrac{6}{13}
C1213\dfrac{12}{13}
D1312\dfrac{13}{12}correct
Solution
Step 1: This is a linear ODE dydx+P(x) y=Q(x)\dfrac{dy}{dx}+P(x)\,y=Q(x) with Q(x)=2+e−2xQ(x)=2+e^{-2x}. Split P(x)P(x) by first writing the 3x2x3+2\dfrac{3x^2}{x^3+2} piece over the common denominator:
3x2x3+2=3x2(2+e−2x)(x3+2)(2+e−2x)=6x2+3x2e−2x(x3+2)(2+e−2x).\dfrac{3x^2}{x^3+2}=\dfrac{3x^2\left(2+e^{-2x}\right)}{(x^3+2)(2+e^{-2x})}=\dfrac{6x^2+3x^2e^{-2x}}{(x^3+2)(2+e^{-2x})}.
 Step 2: Subtract this from P(x)P(x):
P(x)−3x2x3+2=(2x3+4)e−2x(x3+2)(2+e−2x)=2(x3+2)e−2x(x3+2)(2+e−2x)=2e−2x2+e−2x.P(x)-\dfrac{3x^2}{x^3+2}=\dfrac{\left(2x^3+4\right)e^{-2x}}{(x^3+2)(2+e^{-2x})}=\dfrac{2(x^3+2)e^{-2x}}{(x^3+2)(2+e^{-2x})}=\dfrac{2e^{-2x}}{2+e^{-2x}}.
∴ P(x)=3x2x3+2+2e−2x2+e−2x.\therefore\ P(x)=\dfrac{3x^2}{x^3+2}+\dfrac{2e^{-2x}}{2+e^{-2x}}.
 Step 3: Find the integrating factor. Since ddxln⁡(x3+2)=3x2x3+2\dfrac{d}{dx}\ln(x^3+2)=\dfrac{3x^2}{x^3+2} and ddxln⁡(2+e−2x)=−2e−2x2+e−2x\dfrac{d}{dx}\ln\left(2+e^{-2x}\right)=\dfrac{-2e^{-2x}}{2+e^{-2x}},
∫P dx=ln⁡(x3+2)−ln⁡(2+e−2x)=ln⁡ ⁣x3+22+e−2x,I.F.=x3+22+e−2x.\int P\,dx=\ln(x^3+2)-\ln\left(2+e^{-2x}\right)=\ln\!\dfrac{x^3+2}{2+e^{-2x}},\qquad \text{I.F.}=\dfrac{x^3+2}{2+e^{-2x}}.
 Step 4: Multiply through by the I.F. The right side collapses:
ddx ⁣(y⋅x3+22+e−2x)=(2+e−2x)⋅x3+22+e−2x=x3+2.\dfrac{d}{dx}\!\left(y\cdot\dfrac{x^3+2}{2+e^{-2x}}\right)=\left(2+e^{-2x}\right)\cdot\dfrac{x^3+2}{2+e^{-2x}}=x^3+2.
 Step 5: Integrate: y⋅x3+22+e−2x=x44+2x+C.y\cdot\dfrac{x^3+2}{2+e^{-2x}}=\dfrac{x^4}{4}+2x+C. Step 6: Apply y(0)=32y(0)=\dfrac32. At x=0x=0: I.F. =22+1=23=\dfrac{2}{2+1}=\dfrac23, so LHS =32⋅23=1=\dfrac32\cdot\dfrac23=1; RHS =C=C. Hence C=1C=1, giving
y=(2+e−2x)x3+2(x44+2x+1).y=\dfrac{\left(2+e^{-2x}\right)}{x^3+2}\left(\dfrac{x^4}{4}+2x+1\right).
 Step 7: Evaluate at x=1x=1: x3+2=3x^3+2=3 and x44+2x+1=14+2+1=134\dfrac{x^4}{4}+2x+1=\dfrac14+2+1=\dfrac{13}{4}. So
y(1)=(2+e−2)⋅13/43=1312(2+e−2).y(1)=\left(2+e^{-2}\right)\cdot\dfrac{13/4}{3}=\dfrac{13}{12}\left(2+e^{-2}\right).
∴ α=1312.\therefore\ \alpha=\dfrac{13}{12}.
 Correct answer: α=1312\alpha=\dfrac{13}{12}
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