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

$\begin{aligned} y \cdot \log x = \int \log x \cdot 2 d x \\ \Rightarrow y \cdot \log x = 2 \int \log x d x \\ \Rightarrow y \cdot \log x = 2 [x \log x - x] + c \end{aligned}$

[

]

पिछला

अगला

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

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Differential Equations 2 Question 16

16. Let y(x) be the solution of the differential equation (xlog⁡x)dydx+y=2xlog⁡x,(x≥1). Then, y(e) is equal to

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Menu

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