
The differential equation found by the elimination of the arbitrary constant \[{\rm{K}}\] from the equation\[y = (x + K){e^{ - x}}\]is
A. \[\frac{{dy}}{{dx}} - y = {e^{ - x}}\]
B. \[\frac{{dy}}{{dx}} - y{e^x} = 1\]
C. \[\frac{{dy}}{{dx}} + y{e^x} = 1\]
D. \[\frac{{dy}}{{dx}} + y = {e^{ - x}}\]
Answer
163.5k+ views
Hint:
Before attempting to answer the issue, it is important to understand what a differential equation is and what the arbitrary constants are in all equations. Following that, we will compare y to x and discriminate between them. From this, we shall have an equation in the form of x, y, and y's derivative. We shall once more distinguish the resulting equation with regard to x. The second derivative of \[y,x\], and \[y\] will be given by this equation as well. We will then calculate the relationship between\[y\]and\[y\]'s second derivative from this.
Formula used:
\[y = {x^n}\]
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = n{x^{n-1}}\]
\[y = {e^{-x}}\]
\[\frac{{dy}}{{dx}} ={-e^{-x}}\]
Complete step-by-step solution:
We have been given the equation of the arbitrary constant \[{\rm{K}}\]
\[y = (x +K){e^{ - x}}............(1)\]
On differentiating the equation (1) with respect to \[x\], we obtain
\[\frac{{dy}}{{dx}} = {e^{ - x}}[1] + (x +K){e^{ - x}}( - 1)\]
Simplify the above equation to make it less complicated:
\[ \Rightarrow \frac{{dy}}{{dx}} = {e^{ - x}} - y...........(2)\]
Let’s rearrange the equation (2) explicitly to have \[{e^{ - x}}\]on one side, we get
\[ \Rightarrow \frac{{dy}}{{dx}} + y = {e^{ - x}}\]
Therefore, the differential equation found by the elimination of the arbitrary constant \[{\rm{K}}\] from the equation \[y = (x + K){e^{ - x}}\] is \[\frac{{dy}}{{dx}} + y = {e^{ - x}}\]
Hence, the option D is correct.
Note:
Eliminating the arbitrary constants from the equation before forming a differential equation is a need. In general, we differentiate the given equation so many times that the number of arbitrary constants is equal to the number of times. The idea is that we need to remove the arbitrary constant from the equation, not that we need to discover the differential equation. The differentiation may vary the same amount of times.
Before attempting to answer the issue, it is important to understand what a differential equation is and what the arbitrary constants are in all equations. Following that, we will compare y to x and discriminate between them. From this, we shall have an equation in the form of x, y, and y's derivative. We shall once more distinguish the resulting equation with regard to x. The second derivative of \[y,x\], and \[y\] will be given by this equation as well. We will then calculate the relationship between\[y\]and\[y\]'s second derivative from this.
Formula used:
\[y = {x^n}\]
Differentiate with respect to x
\[\frac{{dy}}{{dx}} = n{x^{n-1}}\]
\[y = {e^{-x}}\]
\[\frac{{dy}}{{dx}} ={-e^{-x}}\]
Complete step-by-step solution:
We have been given the equation of the arbitrary constant \[{\rm{K}}\]
\[y = (x +K){e^{ - x}}............(1)\]
On differentiating the equation (1) with respect to \[x\], we obtain
\[\frac{{dy}}{{dx}} = {e^{ - x}}[1] + (x +K){e^{ - x}}( - 1)\]
Simplify the above equation to make it less complicated:
\[ \Rightarrow \frac{{dy}}{{dx}} = {e^{ - x}} - y...........(2)\]
Let’s rearrange the equation (2) explicitly to have \[{e^{ - x}}\]on one side, we get
\[ \Rightarrow \frac{{dy}}{{dx}} + y = {e^{ - x}}\]
Therefore, the differential equation found by the elimination of the arbitrary constant \[{\rm{K}}\] from the equation \[y = (x + K){e^{ - x}}\] is \[\frac{{dy}}{{dx}} + y = {e^{ - x}}\]
Hence, the option D is correct.
Note:
Eliminating the arbitrary constants from the equation before forming a differential equation is a need. In general, we differentiate the given equation so many times that the number of arbitrary constants is equal to the number of times. The idea is that we need to remove the arbitrary constant from the equation, not that we need to discover the differential equation. The differentiation may vary the same amount of times.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main Chemistry Question Paper with Answer Keys and Solutions

JEE Main Reservation Criteria 2025: SC, ST, EWS, and PwD Candidates

What is Normality in Chemistry?

Chemistry Electronic Configuration of D Block Elements: JEE Main 2025

Other Pages
NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks
