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

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  • $y = vx$
  • $\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

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