SATHEE: Differential Equations 2 Question 16

Differential Equations 2 Question 16

16.

Let $y (x)$ be the solution of the differential equation $(x \log x) \frac{d y}{d x} + y = 2 x \log x, (x \geq 1)$ . Then, $y (e)$ is equal to

(a) e

(b) 0

(d) $2 e$

(2015 Main)

Show Answer

Answer:

Correct Answer: 16. (c)

Solution:

Given differential equation is

$\begin{aligned} (x \log x) \frac{d y}{d x} + y = 2 x \log x \\ \Rightarrow \frac{d y}{d x} + \frac{y}{x \log x} = 2 \end{aligned}$

This is a linear differential equation.

$∴ I F = e^{\int \frac{1}{x \log x} d x} = e^{\log (\log x)} = \log x$

Now, the solution of given differential equation is given by

$y \cdot \log x = \int \log x \cdot 2 d x$

$\Rightarrow y \cdot \log x = 2 \int \log x d x$

$\Rightarrow y \cdot \log x = 2 [x \log x - x] + c$

At $x = 1 \Rightarrow c = 2$

$\Rightarrow y . l o g x = 2 [x l o g x - x] + 2$

At $x = e, y = 2 (e - e) + 2$

$y = 2$

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Differential Equations 2 Question 16

16.

Answer:

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