Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

Solve the following differential equation : \( (x^2-y^2)dx+2xydy=0\) given that y = 1 when x = 1.

Can you answer this question?

1 Answer

0 votes
  • $y = vx$
  • $\large\frac{dy}{dx }$$= v + x\large\frac{dv}{dx}$
Step 1:
By rearranging the above equation we get,
$\large\frac{dy}{dx} = -\large\frac{(x^2-y^2)}{2xy }$
Let us substitute $y = vx$
$v + x\large\frac{dv}{dx} = -\large\frac{(x^2-v^2x^2)}{2xvx}$
Step 2:
Taking $x^2$ as the common factor and cancelling it on the RHS
$v + x\large\frac{dv}{dx} = \frac{(1-v^2)}{2v}$
Bringing $v$ from LHS to the RHS we get,
$x\large\frac{dv}{dx }= -\frac{(1- v^2) + 2v^2}{2v}$
$x\large\frac{dv}{dx} = -\frac{(1+v^2)}{2v}$
On seperating the variables we get
$2v\large\frac{dv}{(1+v^2) }= -\frac{ dx}{x}$
Step 3:
On integrating both sides we get
$\int \large\frac{2vdv}{(1+v^2)} = - \int\large\frac{ dx}{x }$
$\log(1+v^2) = -\log x +\log C$
$\log\large\frac{(1+v^2)}{x} $$= \log C$
$\large\frac{(1+v^2)}{x} $$= \log C$
Step 4:
Writing the value of $v = \large\frac{y}{x}$ we get
$\large\frac{[1+(y^2/x^2)]}{x}$$ = C$
$x^2+y^2 = Cx$
This is the required solution.
Step 5:
When $x=1,y=1$
$1+1=C\times 1$
So the required solution is $x^2+y^2=2x$
answered Sep 26, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App