
Find the number of solutions of the differential equation \[y' = \dfrac{{y + 1}}{{x - 1}}\], if \[y\left( 1 \right) = 2\].
A. None
B. One
C. Two
D. Infinite
Answer
161.1k+ views
Hint: Here, a first-order differential equation is given. First, rewrite \[y'\] as \[\dfrac{{dy}}{{dx}}\] and simplify the differential equation. Then, integrate both sides with respect to the corresponding variables. After that, solve the integrals and find the solution of the differential equation. Then, substitute the values from the given equation \[y\left( 1 \right) = 2\] in the solution of the differential equation and find the value of the integration constant \[c\]. In the end, verify the conditions and find the required answer.
Formula Used: \[\int {\dfrac{1}{x}} dx = \log x + c\]
\[\log a + \log b = \log \left( {ab} \right)\]
Complete step by step solution: Given:
The first-order differential equation is \[y' = \dfrac{{y + 1}}{{x - 1}}\] and \[y\left( 1 \right) = 2\].
Let’s solve the differential equation.
Rewrite \[y'\] as \[\dfrac{{dy}}{{dx}}\].
\[\dfrac{{dy}}{{dx}} = \dfrac{{y + 1}}{{x - 1}}\]
Simplify the equation.
\[\dfrac{{dy}}{{y + 1}} = \dfrac{{dx}}{{x - 1}}\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\dfrac{{dy}}{{y + 1}}} = \int {\dfrac{{dx}}{{x - 1}}} \]
Apply the integration formulas \[\int {\dfrac{1}{x}} dx = \log x + c\].
\[\log \left( {y + 1} \right) = \log \left( {x - 1} \right) + \log c\]
Apply the product property of the logarithm on the right-hand side.
\[\log \left( {y + 1} \right) = \log \left[ {\left( {x - 1} \right)c} \right]\]
Equate both sides.
\[\left( {y + 1} \right) = \left( {x - 1} \right)c\]
\[ \Rightarrow \dfrac{{y + 1}}{{x - 1}} = c\] \[.....\left( 1 \right)\]
Now to calculate the value of the integration constant, substitute the values of the given equation \[y\left( 1 \right) = 2\] in the equation \[\left( 1 \right)\].
We have, \[x = 1\] and \[y = 2\]
Then,
\[\dfrac{{2 + 1}}{{1 - 1}} = c\]
\[ \Rightarrow \dfrac{3}{0} = c\]
\[ \Rightarrow c = 0\]
So, the required solution is possible only when \[x - 1 = 0\].
Therefore, only one solution exists for the given differential equation \[y' = \dfrac{{y + 1}}{{x - 1}}\].
Option ‘B’ is correct
Note: Students often apply a wrong formula to integrate \[\dfrac{1}{x}\] . They integrate it by using the power formula of integration. But the correct formula is \[\int {\dfrac{1}{x}dx} = \log x + c\].
Formula Used: \[\int {\dfrac{1}{x}} dx = \log x + c\]
\[\log a + \log b = \log \left( {ab} \right)\]
Complete step by step solution: Given:
The first-order differential equation is \[y' = \dfrac{{y + 1}}{{x - 1}}\] and \[y\left( 1 \right) = 2\].
Let’s solve the differential equation.
Rewrite \[y'\] as \[\dfrac{{dy}}{{dx}}\].
\[\dfrac{{dy}}{{dx}} = \dfrac{{y + 1}}{{x - 1}}\]
Simplify the equation.
\[\dfrac{{dy}}{{y + 1}} = \dfrac{{dx}}{{x - 1}}\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\dfrac{{dy}}{{y + 1}}} = \int {\dfrac{{dx}}{{x - 1}}} \]
Apply the integration formulas \[\int {\dfrac{1}{x}} dx = \log x + c\].
\[\log \left( {y + 1} \right) = \log \left( {x - 1} \right) + \log c\]
Apply the product property of the logarithm on the right-hand side.
\[\log \left( {y + 1} \right) = \log \left[ {\left( {x - 1} \right)c} \right]\]
Equate both sides.
\[\left( {y + 1} \right) = \left( {x - 1} \right)c\]
\[ \Rightarrow \dfrac{{y + 1}}{{x - 1}} = c\] \[.....\left( 1 \right)\]
Now to calculate the value of the integration constant, substitute the values of the given equation \[y\left( 1 \right) = 2\] in the equation \[\left( 1 \right)\].
We have, \[x = 1\] and \[y = 2\]
Then,
\[\dfrac{{2 + 1}}{{1 - 1}} = c\]
\[ \Rightarrow \dfrac{3}{0} = c\]
\[ \Rightarrow c = 0\]
So, the required solution is possible only when \[x - 1 = 0\].
Therefore, only one solution exists for the given differential equation \[y' = \dfrac{{y + 1}}{{x - 1}}\].
Option ‘B’ is correct
Note: Students often apply a wrong formula to integrate \[\dfrac{1}{x}\] . They integrate it by using the power formula of integration. But the correct formula is \[\int {\dfrac{1}{x}dx} = \log x + c\].
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
