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Mathematics · Differential Equations

JEE Main 2024 — 8 April, Shift 1 — Question 6

Let f(x)f(x) be a positive function such that the area bounded by y=f(x),y=0y=f(x), y=0 from x=0x=0 to x=a>0x=a>0 is e−a+4a2+a−1\mathrm{e}^{-\mathrm{a}}+4 \mathrm{a}^{2}+\mathrm{a}-1. Then the differential equation, whose general solution is y=c1f(x)+c2\mathrm{y}=\mathrm{c}_{1} f(\mathrm{x})+\mathrm{c}_{2}, where c1\mathrm{c}_{1} and c2\mathrm{c}_{2} are arbitrary constants, is :

  1. Option A:

    (8ex−1)d2ydx2+dydx=0\left(8 e^{x}-1\right) \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0

  2. Option B:

    (8ex+1)d2ydx2−dydx=0\left(8 e^{x}+1\right) \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0

  3. Option C:

    (8ex+1)d2ydx2+dydx=0\left(8 e^{x}+1\right) \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0

    Correct
  4. Option D:

    (8ex−1)d2ydx2−dydx=0\left(8 e^{x}-1\right) \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0

Answer: C

Step-by-step solution

∫0af(x)dx=e−a+4a2+a−1\int_{0}^{a} f(x) d x=e^{-a}+4 a^{2}+a-1

f(a)=−e−a+8a+1f(a)=-e^{-a}+8 a+1

f(x)=−e−x+8x+1f(x)=-e^{-x}+8 x+1

Now y=C1f(x)+C2y=C_{1} \mathrm{f}(\mathrm{x})+\mathrm{C}_{2}

dydx=C1f′(x)=C1(e−x+8)\frac{d y}{d x}=C_{1} f^{\prime}(x)=C_{1}\left(e^{-x}+8\right)

d2ydx2=−C1e−x⇒−exd2ydx2\frac{d^{2} y}{d x^{2}}=-C_{1} e^{-x} \Rightarrow-e^{x} \frac{d^{2} y}{d x^{2}}

Put in equation (1)

dydx=−exd2ydx2(e−x+8)\frac{d y}{d x}=-e^{x} \frac{d^{2} y}{d x^{2}}\left(e^{-x}+8\right)

(8ex+1)d2ydx2+dydx=0\left(8 e^{x}+1\right) \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 1
Subject
Mathematics
Chapter
Differential Equations
Topic
Formation Of D.E

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