$\displaystyle

y - x\frac{dy}{dx} = \sqrt{1 + (\frac{dy}{dx})^2}(x^2 + y^2)

$

I Have Tried It At my level best but was unable to solve it!

Printable View

- Aug 28th 2010, 06:53 PMraj007Find General Solution?
$\displaystyle

y - x\frac{dy}{dx} = \sqrt{1 + (\frac{dy}{dx})^2}(x^2 + y^2)

$

I Have Tried It At my level best but was unable to solve it! - Aug 28th 2010, 07:02 PMProve It
$\displaystyle y - x\,\frac{dy}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,(x^2 + y^2)$

$\displaystyle \left(y - x\,\frac{dy}{dx}\right)^2 = \left[1 + \left(\frac{dy}{dx}\right)^2\right](x^2 + y^2)^2$

$\displaystyle y^2 - 2xy\,\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = \left[1 + \left(\frac{dy}{dx}\right)^2\right](x^4 + 2x^2y^2 + y^4)$

$\displaystyle y^2 - 2xy\,\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = x^4 + 2x^2y^2 + y^4 + (x^4 + 2x^2y^2 + y^4)\left(\frac{dy}{dx}\right)^2$

$\displaystyle (x^4 + 2x^2y^2 + y^4 - 1)\left(\frac{dy}{dx}\right)^2 + 2xy\,\frac{dy}{dx} + x^4 + 2x^2y^2 + y^4 - y^2 = 0$.

This is now a Quadratic in $\displaystyle \frac{dy}{dx}$, so solve using the Quadratic Formula. - Aug 28th 2010, 07:16 PMraj007
the solution of dy/sx dont seems to be integrate able at all.

i dont think this is gonna work!

btw these question are from section where we have to use substitutions like

x = r.cost & y = r.sint - Aug 28th 2010, 07:58 PMProve It
Just use the Quadratic formula. It's an equation of the form

$\displaystyle aX^2 + bX + c = 0$.

Here $\displaystyle X = \frac{dy}{dx}, a = x^4 + 2x^2y^2 + y^4 -1, b = 2xy, c = x^4 + 2x^2y^2 + y^4 - y^2$.

So the solutions are

$\displaystyle \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

It's ok to keep your solution in terms of $\displaystyle x$ and $\displaystyle y$. You don't need to do any integration at all. - Aug 29th 2010, 05:29 AMJester
- Aug 29th 2010, 08:20 AMraj007
can u show me how u got

$\displaystyle r_t^2+r^2-1 = 0$