T ttn181 New member Joined Mar 30, 2011 Messages 3 Apr 15, 2011 #1 How do I find the general solution to this differential equation? I just don't know where to start. :/ Attachments Capture.JPG 9.7 KB · Views: 76
How do I find the general solution to this differential equation? I just don't know where to start. :/
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Apr 21, 2011 #2 It's been a while. Hope it is not too late. I just noticed your problem. Anyway, this looks like a Bernoulli Equation if we rearrange: \(\displaystyle xdy=x^{m}y^{n}dx-ydx\) \(\displaystyle xdy=(x^{m}y^{n}-y)dx\) Separate variables: \(\displaystyle \frac{dy}{dx}=\frac{x^{m}y^{n}-y}{x}\) \(\displaystyle \frac{dy}{dx}+\frac{y}{x}=x^{m-1}y^{n}\) The general solution would then be: \(\displaystyle y=\frac{1}{\left(\frac{x^{m}(n-1)}{n-m-1}+C_{1}x^{n-1}\right)^{\frac{1}{n-1}}}\)
It's been a while. Hope it is not too late. I just noticed your problem. Anyway, this looks like a Bernoulli Equation if we rearrange: \(\displaystyle xdy=x^{m}y^{n}dx-ydx\) \(\displaystyle xdy=(x^{m}y^{n}-y)dx\) Separate variables: \(\displaystyle \frac{dy}{dx}=\frac{x^{m}y^{n}-y}{x}\) \(\displaystyle \frac{dy}{dx}+\frac{y}{x}=x^{m-1}y^{n}\) The general solution would then be: \(\displaystyle y=\frac{1}{\left(\frac{x^{m}(n-1)}{n-m-1}+C_{1}x^{n-1}\right)^{\frac{1}{n-1}}}\)
