Let f be a differential function with ₓ→∞}f(x)=0. If y' + y f'(x) - f(x) f'(x) = 0, ₓ→∞} y(x) = 0 then

Correct Answer is : y+1=e−f(x)+f(x)y+1 = e^{-f(x)} + f(x)y+1=e−f(x)+f(x)

Correct Answer is : y+1=e−f(x)+f(x)y+1 = e^{-f(x)} + f(x)y+1=e−f(x)+f(x)

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WBJEE 2024 Math Paper

Question : 3 of 75

Marks: +1, -0

\lim\limits_{x\to\infty}f(x)=0

y' + y f'(x) - f(x) f'(x) = 0, \lim\limits_{x\to\infty} y(x) = 0

then

Solution:

\lim\limits_{x\to\infty}f(x)=0

y' + y f'(x) - f(x) \cdot f'(x) = 0

IF = e^{\int f'(x) \, dx} = e^{f(x)}

y \cdot e^{f(x)} = \int e^{f(x)} \cdot f(x) \cdot f'(x) \, dx

\text{ let } f(x)=u, f'(x)\,dx=du

\therefore y e^{f(x)} = \int e^u u \, du

\Rightarrow y e^{f(x)} = e^u (u-1) + c \\ y e^{f(x)} = e^{f(x)} (f(x)-1) + c

Now

\lim\limits_{x\to\infty}f(x)=0

\therefore \text{ at } x\to\infty, f(x)=y=0

\Rightarrow 0 \cdot e^0 = e^0 (0-1) + c = 0

c=1

\text{ so, } y e^{f(x)} = e^{f(x)} \cdot (f(x)-1) + 1

\Rightarrow y = f(x) - 1 + \frac{1}{e^{f(x)}}

\Rightarrow y+1 = f(x) + e^{-f(x)}

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