I am trying to solve this differential equation and I can't get started and was wondering if I could get some help with a kick start:

$\displaystyle (\frac{dy}{dx})^2 +(\frac{dy}{dx})*x-\frac{dy}{dx}-x=0$

Any help would be greatly appreciated!

MT

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- Feb 2nd 2011, 02:12 PMEmptyfirst order non-linear differential equation
I am trying to solve this differential equation and I can't get started and was wondering if I could get some help with a kick start:

$\displaystyle (\frac{dy}{dx})^2 +(\frac{dy}{dx})*x-\frac{dy}{dx}-x=0$

Any help would be greatly appreciated!

MT - Feb 2nd 2011, 02:23 PMAckbeet
I know it sounds crazy, but couldn't you use the quadratic formula here to solve for y'? That is, you have

$\displaystyle (y')^{2}+y'(x-1)-x=0,$ so

$\displaystyle y'=\dfrac{-(x-1)\pm\sqrt{(x-1)^{2}-4(-x)}}{2}\dots$

Does that work? - Feb 2nd 2011, 02:23 PMTheEmptySet
Notice that

$\displaystyle \displaystyle \left(\frac{dy}{dx}\right)^2 +\left(\frac{dy}{dx}\right)x-\frac{dy}{dx}-x=0 \iff \left(\frac{dy}{dx}\right)^2 -\frac{dy}{dx}+\left(\frac{dy}{dx}\right)x-x=0 $

factoring gives

$\displaystyle \displaystyle \frac{dy}{dx}\left( \frac{dy}{dx}-1\right)+x\left( \frac{dy}{dx}-1\right)=0 $

$\displaystyle \displaystyle \left( \frac{dy}{dx}-1\right)\left( \frac{dy}{dx}+x\right)=0$

Use the zero product principle. This will give two different solutions. Remember for nonlinear equations that the principle of superposition does not hold. So you cannot add the solutions to obtain a more general solution. - Feb 2nd 2011, 02:50 PMAckbeet
Reply to TheEmptySet:

Ah, your method is more elegant, because you don't get those annoying magnitude signs as follows:

$\displaystyle y'=\dfrac{-x+1\pm|x+1|}{2}.$ - Feb 2nd 2011, 03:02 PMEmpty
Thank you all for your help... I should have known to use the quadratic equation!! Thanks TheEmptySet for that alternative approach!