JAMIA 2023 — Mathematics PYQ

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Question

JAMIA 2023 — Mathematics PYQ

JAMIA | Mathematics | 2023

Solution of the differential equation $\frac{d x}{d y} - \frac{x lo g x}{1 + lo g x} = \frac{e ^{y}}{1 + lo g x}$ , if $y (1) = 0$ , is

Choose the correct answer:

Explanation

. Given Differential Equation:

\frac{dx}{dy} - \frac{x \log x}{1 + \log x} = \frac{e^y}{1 + \log x}

Multiply by $(1 + \log x)$ :

(1 + \log x) \frac{dx}{dy} - x \log x = e^y

2. Substitution:

Let $u = x \log x$

Differentiating with respect to $y$ :

\frac{du}{dy} = (1 + \log x) \frac{dx}{dy}

3. Linear Equation solve karein:

\frac{du}{dy} - u = e^y

I.F. = e^{\int -1 dy} = e^{-y}

Solution:

u \cdot e^{-y} = \int e^y \cdot e^{-y} \, dy + C

u \cdot e^{-y} = y + C

x \log x = (y + C)e^y

4. Initial Condition ( $y(1) = 0$ ) apply karein:

1 \cdot \log 1 = (0 + C)e^0

0 = C

5. Result:

x \log x = y e^y

Dono taraf exponential ( $e$ ) lene par:

e^{x \log x} = e^{y e^y}

x^x = e^{y e^y}