
What is the solution of the differential equation \[x \dfrac{{dy}}{{dx}} + y = {y^2}\]?
A. \[y = 1 + cxy\]
B. \[y = \log\left( {cxy} \right)\]
C. \[y + 1 = cxy\]
D. \[y = c + xy\]
Answer
162k+ views
Hint: Here, the first order differential equation is given. First, simplify the given equation. Then, integrate both sides of the equation with respect to the corresponding variables. After that, solve both integrals by using the standard integral formulas. In the end, apply the property of a logarithm to get the solution of the differential equation.
Formula used:
\[\int {\dfrac{{dx}}{x}} = \log\left( x \right)\]
The logarithmic properties:
\[\log\left( a \right) - \log\left( b \right) = \log\left( {\dfrac{a}{b}} \right)\]
\[\log\left( a \right) + \log\left( b \right) = \log\left( {ab} \right)\]
Complete step by step solution:
The given differential equation is \[x \dfrac{{dy}}{{dx}} + y = {y^2}\].
Let’s find out the solution of the given differential equation.
Simplify the given equation.
\[x \dfrac{{dy}}{{dx}} = {y^2} - y\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^2} - y}}{x}\]
\[ \Rightarrow \dfrac{{dy}}{{{y^2} - y}} = \dfrac{{dx}}{x}\]
Simplify the left-hand side of the above equation.
\[\left[ {\dfrac{1}{{y - 1}} - \dfrac{1}{y}} \right]dy = \dfrac{{dx}}{x}\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\left[ {\dfrac{1}{{y - 1}} - \dfrac{1}{y}} \right]dy = } \int {\dfrac{{dx}}{x}} \]
Apply the integration rule on the left-hand side.
\[\int {\dfrac{1}{{y - 1}}} dy - \int {\dfrac{1}{y}} dy = \int {\dfrac{{dx}}{x}} \]
Solve the integrals by using the standard formula \[\int {\dfrac{{dx}}{x}} = \log\left( x \right)\].
\[\log\left( {y - 1} \right) - \log y = \log x + \log c\]
Now apply the logarithmic properties.
We get,
\[\log\left( {\dfrac{{y - 1}}{y}} \right) = \log \left( {cx} \right)\]
Equate both sides.
\[\dfrac{{y - 1}}{y} = cx\]
Solve the above equation.
\[y - 1 = cxy\]
\[ \Rightarrow y = 1 + cxy\]
Hence the correct option is A.
Note: It is necessary to use an integration constant as soon as integration is performed if we solve a first-order differential equation by the variable separable method.
If all terms of the solution are in the form of a logarithm, then the integration constant is also in the form of a logarithm.
Formula used:
\[\int {\dfrac{{dx}}{x}} = \log\left( x \right)\]
The logarithmic properties:
\[\log\left( a \right) - \log\left( b \right) = \log\left( {\dfrac{a}{b}} \right)\]
\[\log\left( a \right) + \log\left( b \right) = \log\left( {ab} \right)\]
Complete step by step solution:
The given differential equation is \[x \dfrac{{dy}}{{dx}} + y = {y^2}\].
Let’s find out the solution of the given differential equation.
Simplify the given equation.
\[x \dfrac{{dy}}{{dx}} = {y^2} - y\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^2} - y}}{x}\]
\[ \Rightarrow \dfrac{{dy}}{{{y^2} - y}} = \dfrac{{dx}}{x}\]
Simplify the left-hand side of the above equation.
\[\left[ {\dfrac{1}{{y - 1}} - \dfrac{1}{y}} \right]dy = \dfrac{{dx}}{x}\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\left[ {\dfrac{1}{{y - 1}} - \dfrac{1}{y}} \right]dy = } \int {\dfrac{{dx}}{x}} \]
Apply the integration rule on the left-hand side.
\[\int {\dfrac{1}{{y - 1}}} dy - \int {\dfrac{1}{y}} dy = \int {\dfrac{{dx}}{x}} \]
Solve the integrals by using the standard formula \[\int {\dfrac{{dx}}{x}} = \log\left( x \right)\].
\[\log\left( {y - 1} \right) - \log y = \log x + \log c\]
Now apply the logarithmic properties.
We get,
\[\log\left( {\dfrac{{y - 1}}{y}} \right) = \log \left( {cx} \right)\]
Equate both sides.
\[\dfrac{{y - 1}}{y} = cx\]
Solve the above equation.
\[y - 1 = cxy\]
\[ \Rightarrow y = 1 + cxy\]
Hence the correct option is A.
Note: It is necessary to use an integration constant as soon as integration is performed if we solve a first-order differential equation by the variable separable method.
If all terms of the solution are in the form of a logarithm, then the integration constant is also in the form of a logarithm.
Recently Updated Pages
If tan 1y tan 1x + tan 1left frac2x1 x2 right where x frac1sqrt 3 Then the value of y is

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 Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

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

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

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

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

1 Billion in Rupees

Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE
