Jul 18, 2013 Thread starter Admin #1 M MarkFL Administrator Staff member Feb 24, 2012 13,775 Find the general solution for: \(\displaystyle \left(\left(x^2+y^2 \right)x-y \right)\,dx+\left(\left(x^2+y^2 \right)y+x \right)\,dy=0\)
Find the general solution for: \(\displaystyle \left(\left(x^2+y^2 \right)x-y \right)\,dx+\left(\left(x^2+y^2 \right)y+x \right)\,dy=0\)
Jul 19, 2013 #2 Jester Well-known member MHB Math Helper Jan 26, 2012 183 My solution Spoiler If you switch to polar coords $x = r \cos \theta , \; y = r \sin \theta$ the ODE becomes $r dr + dt = 0$. At this point the ODE is trivial.
My solution Spoiler If you switch to polar coords $x = r \cos \theta , \; y = r \sin \theta$ the ODE becomes $r dr + dt = 0$. At this point the ODE is trivial.
Jul 19, 2013 Thread starter Admin #3 M MarkFL Administrator Staff member Feb 24, 2012 13,775 Jester said: My solution Spoiler If you switch to polar coords $x = r \cos \theta , \; y = r \sin \theta$ the ODE becomes $r dr + dt = 0$. At this point the ODE is trivial. Click to expand... Brilliant! For those who may not follow the substitution and subsequent result, I will elaborate, and also give my method here: Spoiler Jester suggests switching to polar coordinates: \(\displaystyle x=r\cos(\theta)\) \(\displaystyle y=r\sin(\theta)\) and so we find: \(\displaystyle x^2+y^2=r^2\) \(\displaystyle \theta=\tan^{-1}\left(\frac{y}{x} \right)\) \(\displaystyle dx=-r\sin(\theta)\,d\theta+\cos(\theta)\,dr\) \(\displaystyle dy=r\cos(\theta)\,d\theta+\sin(\theta)\,dr\) Now, substituting into the ODE, we get: \(\displaystyle \left(r^3\cos(\theta)- r\sin(\theta) \right)\left(\cos(\theta)\,dr- r\sin(\theta)\,d\theta \right)+ \left(r^3\sin(\theta)+ r\cos(\theta) \right)\left(\sin(\theta)\,dr+ r\cos(\theta)\,d\theta \right)=0\) Dividing through by $r$ and expanding, we find the first term is: \(\displaystyle r^2\cos^2(\theta)\,dr-r^3\sin(\theta)\cos(\theta)-\sin(\theta)\cos(\theta)\,dr+r\sin^2(\theta)\,dt\) and the second term is: \(\displaystyle r^2\sin^2(\theta)\,dr+r^3\sin(\theta)\cos(\theta)+\sin(\theta)\cos(\theta)\,dr+r\cos^2(\theta)\,dt\) And so their sum is (and applying the Pythagorean identity): \(\displaystyle r^2\,dr+r\,d\theta=0\) Divide through by $r$ to obtain: \(\displaystyle r\,dr+d\theta=0\) Integrating, we find: \(\displaystyle r^2+2\theta=C\) And back-substituting, we get the general solution: \(\displaystyle x^2+y^2+2\tan^{-1}\left(\frac{y}{x} \right)=C\) This is the method I used: Beginning with: \(\displaystyle \left(\left(x^2+y^2 \right)x-y \right)\,dx+\left(\left(x^2+y^2 \right)y+x \right)\,dy=0\) \(\displaystyle \left(x^2+y^2 \right)x-y+\left(\left(x^2+y^2 \right)y+x \right)y'=0\) Divide through by \(\displaystyle x^2+y^2\) to get: \(\displaystyle x+yy'+\frac{xy'-y}{x^2+y^2}=0\) \(\displaystyle 2x+2yy'+2\frac{1}{\left(\frac{y}{x} \right)^2+1}\cdot\frac{xy'-y}{x^2}=0\) \(\displaystyle 2x+2y\frac{dy}{dx}+2\frac{1}{\left(\frac{y}{x} \right)^2+1}\frac{d}{dx}\left(\frac{y}{x} \right)=0\) Integrating with respect to $x$, we obtain the general solution: \(\displaystyle x^2+y^2+2\tan^{-1}\left(\frac{y}{x} \right)=C\)
Jester said: My solution Spoiler If you switch to polar coords $x = r \cos \theta , \; y = r \sin \theta$ the ODE becomes $r dr + dt = 0$. At this point the ODE is trivial. Click to expand... Brilliant! For those who may not follow the substitution and subsequent result, I will elaborate, and also give my method here: Spoiler Jester suggests switching to polar coordinates: \(\displaystyle x=r\cos(\theta)\) \(\displaystyle y=r\sin(\theta)\) and so we find: \(\displaystyle x^2+y^2=r^2\) \(\displaystyle \theta=\tan^{-1}\left(\frac{y}{x} \right)\) \(\displaystyle dx=-r\sin(\theta)\,d\theta+\cos(\theta)\,dr\) \(\displaystyle dy=r\cos(\theta)\,d\theta+\sin(\theta)\,dr\) Now, substituting into the ODE, we get: \(\displaystyle \left(r^3\cos(\theta)- r\sin(\theta) \right)\left(\cos(\theta)\,dr- r\sin(\theta)\,d\theta \right)+ \left(r^3\sin(\theta)+ r\cos(\theta) \right)\left(\sin(\theta)\,dr+ r\cos(\theta)\,d\theta \right)=0\) Dividing through by $r$ and expanding, we find the first term is: \(\displaystyle r^2\cos^2(\theta)\,dr-r^3\sin(\theta)\cos(\theta)-\sin(\theta)\cos(\theta)\,dr+r\sin^2(\theta)\,dt\) and the second term is: \(\displaystyle r^2\sin^2(\theta)\,dr+r^3\sin(\theta)\cos(\theta)+\sin(\theta)\cos(\theta)\,dr+r\cos^2(\theta)\,dt\) And so their sum is (and applying the Pythagorean identity): \(\displaystyle r^2\,dr+r\,d\theta=0\) Divide through by $r$ to obtain: \(\displaystyle r\,dr+d\theta=0\) Integrating, we find: \(\displaystyle r^2+2\theta=C\) And back-substituting, we get the general solution: \(\displaystyle x^2+y^2+2\tan^{-1}\left(\frac{y}{x} \right)=C\) This is the method I used: Beginning with: \(\displaystyle \left(\left(x^2+y^2 \right)x-y \right)\,dx+\left(\left(x^2+y^2 \right)y+x \right)\,dy=0\) \(\displaystyle \left(x^2+y^2 \right)x-y+\left(\left(x^2+y^2 \right)y+x \right)y'=0\) Divide through by \(\displaystyle x^2+y^2\) to get: \(\displaystyle x+yy'+\frac{xy'-y}{x^2+y^2}=0\) \(\displaystyle 2x+2yy'+2\frac{1}{\left(\frac{y}{x} \right)^2+1}\cdot\frac{xy'-y}{x^2}=0\) \(\displaystyle 2x+2y\frac{dy}{dx}+2\frac{1}{\left(\frac{y}{x} \right)^2+1}\frac{d}{dx}\left(\frac{y}{x} \right)=0\) Integrating with respect to $x$, we obtain the general solution: \(\displaystyle x^2+y^2+2\tan^{-1}\left(\frac{y}{x} \right)=C\)