# Find General Solution?

• August 28th 2010, 06:53 PM
raj007
Find General Solution?
$
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!
• August 28th 2010, 07:02 PM
Prove It
$y - x\,\frac{dy}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,(x^2 + y^2)$

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

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

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

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

This is now a Quadratic in $\frac{dy}{dx}$, so solve using the Quadratic Formula.
• August 28th 2010, 07:16 PM
raj007
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
• August 28th 2010, 07:58 PM
Prove It
Just use the Quadratic formula. It's an equation of the form

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

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

So the solutions are

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

It's ok to keep your solution in terms of $x$ and $y$. You don't need to do any integration at all.
• August 29th 2010, 05:29 AM
Jester
Quote:

Originally Posted by raj007
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

Then why not make that substitution. In the end I got

$r_t^2+r^2-1 = 0$.
• August 29th 2010, 08:20 AM
raj007
can u show me how u got

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