JEE MAIN Differential Equation Previous Year Questions (PYQs) – Page 10 of 16

$ \lim_{t \to x} \frac{2t f(x) - x^2 f(t)}{-1} = 3 $

$ x^2 f'(x) - 2x f(x) = 3 $

$ \frac{dy}{dx} - \frac{2y}{x} = \frac{3}{x^2} $

I.F. $ = e^{\int \frac{-2}{x} dx} = e^{-2\log x} = \frac{1}{x^2} $

$ \frac{y}{x^2} = \int \frac{3}{x^4} dx $

$ \frac{y}{x^2} = -\frac{1}{x^3} + c \Rightarrow y = cx^2 - \frac{1}{x} $

$ f(1) = 2 = c - 1 \Rightarrow c = 3 $

$ f(x) = 3x^2 - \frac{1}{x} $

$ f(2) = 12 - \frac{1}{2} $

$ 2f(2) = 23 $

$ f(x) = e^x + \int_0^x y f(y),dy + xe^x \int_0^x f(y),dy $ $ f(x) = e^x + A + Bxe^x $ $ A = \int_0^x y f(y),dy = \int_0^x y(A + e^y + Bye^y),dy $ $ A = \frac{A}{2} + 0(-1) + B(e - 1) $ $ \frac{A}{2} + B(1 - e) = 1 $ $ B = \int_0^1 f(y),dy $ $ B = \int_0^1 (e^y + A + Bye^y),dy $ $ B = (e - 1) + A + B(0 - 1) $ $ B = e - 1 + A + B(0 - (-1)) $ $ B = e - 1 + A + B \Rightarrow A = 1 - e $ $ f(x) = e^x + A + Bxe^x $ $ f(0) = 1 + A + B\cdot 0 = 1 + A = 2 - e $ $ e + f(0) = 2 $