SATHEE: Differential Equations 1 Question 17

Differential Equations 1 Question 17

17. If $P (1) = 0$ and $\frac{d P (x)}{d x} > P (x), \forall x \geq 1$ , then prove that $P (x) > 0, \forall x > 1$ .

(2003, 4M)

Show Answer

Answer:

Correct Answer: 17. (d)

Solution:

Given, $P (1) = 0$ and $\frac{d P (x)}{d x} - P (x) > 0, \forall x \geq 1$

On multiplying Eq. (i) by $e^{- x}$ , we get

$\begin{array}{cc} e^{- x} \cdot \frac{d}{d x} P (x) \cdot \frac{d}{d x} e^{- x} > 0 \\ \Rightarrow & \frac{d}{d x} (P (x) \cdot e^{- x}) > 0 \\ \Rightarrow & P (x) \cdot e^{- x} is an increasing function. \\ \Rightarrow & P (x) \cdot e^{- x} > P (1) \cdot e^{- 1}, \forall x \geq 1 \\ \Rightarrow & P (x) > 0, \forall x > 1 [∵ P (1) = 0 and e^{- x} > 0] \end{array}$

Privacy Policy

Powered by Prutor@IITK

sathee@iitk.ac.in

Media coverage of SATHEE

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Differential Equations 1 Question 17

17. If P(1)=0 and dP(x)dx>P(x),∀x≥1, then prove that P(x)>0,∀x>1.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

17. If $P (1) = 0$ and $\frac{d P (x)}{d x} > P (x), \forall x \geq 1$ , then prove that $P (x) > 0, \forall x > 1$ .