Verified Solution • 2025–26 Session moderate

QJEEM24Mae5b4: JEE Main 2024 Mathematics - Class 12 JEEM24Mae5b4 Solution: Let be a positive function such that the area bounded by from…

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>0 is e−a+4a2+a−1e^{-a}+4 a^2+a-1 . Then the differential equation, whose general solution is y=c1f(x)+c2y=c_1 f(x)+c_2 , where c1c_1 and c2c_2 are arbitrary constants, is

Multiple Choice Options

  • 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
  • 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
  • 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
  • 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

Step-by-Step Verified Solution

∫0af(x)dx=e−a+4a2+a−1

Differentiate equation w.r.t. 'a'

amp;f(a)=−e−a+8a+1amp;⇒f(x)=−e−x+8x+1\begin{aligned}& f(a)=-e^{-a}+8 a+1 \\& \Rightarrow f(x)=-e^{-x}+8 x+1\end{aligned}

And y=c1f(x)+c2y=c_1 f(x)+c_2

amp;y=c1(−e−x+8x+1)+c2amp;y′=c1(e−x+8)⇒c1=y′e−x+8\begin{aligned}& y=c_1\left(-e^{-x}+8 x+1\right)+c_2 \\& y^{\prime}=c_1\left(e^{-x}+8\right) \Rightarrow c_1=\frac{y^{\prime}}{e^{-x}+8}\end{aligned}

y′′=−c1e−xy^{\prime \prime}=-c_1 e^{-x}

put value of c1c_1

amp;d2ydx2=−dydx⋅e−x(e−x+8)=dydx(1+8ex)amp;⇒(1+8ex)d2ydx2+dydx=1\begin{aligned}& \frac{d^2 y}{d x^2}=\frac{-\frac{d y}{d x} \cdot e^{-x}}{\left(e^{-x}+8\right)}=\frac{\frac{d y}{d x}}{\left(1+8 e^x\right)} \\& \Rightarrow\left(1+8 e^x\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=1\end{aligned}

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