Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Solve for particular solution. 2xy + y^2 - 2x^2 dy/dx = 0; y = 2 when x = 1

Solve for particular solution.

    Q15.    2xy + y^2 - 2x^2\frac{dy}{dx} = 0 ;\ y = 2\ \textup{when}\ x = 1

Answers (1)

best_answer

The above eq can be written as;

\frac{dy}{dx}=\frac{2xy+y^{2}}{2x^{2}} = F(x,y)
By looking, we can say that it is a homogeneous equation.

Now, to solve substitute y = vx
Differentiating on both sides wrt x
\frac{dy}{dx}= v +x\frac{dv}{dx}
                                 
Substitute this value in equation (i)

\\=v+x\frac{dv}{dx}= \frac{2v+v^{2}}{2}\\ =x\frac{dv}{dx} = v^{2}/2\\ = \frac{2dv}{v^{2}}=\frac{dx}{x}

integrating on both sides, we get;

\\=-2/v=\log \left | x \right |+C\\ =-\frac{2x}{y}=\log \left | x \right |+C.............................(ii)

Now, y = 2 and x =1, we get

C =-1
put this value in equation(ii)

\\=-\frac{2x}{y}=\log \left | x \right |-1\\ \Rightarrow y = \frac{2x}{1- \log x}

Posted by

manish

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Board exams twice a year from 2025, no plan for semesters: MoE
anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question