For some reason of the many calculus texts I gazed upon there is only one that mentions its name or even the techinque.Originally Posted by shubhadeep
OK, the section of the book in which the question lies has the other questions done by substitution but I couldn't do this one.Originally Posted by ThePerfectHacker
Taking $\displaystyle x=r\,\cos\theta,\,y=r\,\sin\theta$ where $\displaystyle \theta$ & $\displaystyle r$ are variables
So, $\displaystyle x^{2}+y^{2}=r^{2}$
and $\displaystyle \tan\theta\,=\,\frac{y}{x}$
On differentiating the above two equations we can substitute $\displaystyle xdy+ydx$ & $\displaystyle xdy-ydx$ in the differential equation.
So does this help?
First, is this equation correct? Usually if there's an "e" in there it's $\displaystyle e^x$ or something.Originally Posted by shubhadeep
Let's try:
$\displaystyle (xy^{2}-\frac{e}{x^{3}})dx-x^{2}ydy=0$
$\displaystyle x^{2}y \frac{dy}{dx}-xy^{2}=-\frac{e}{x^{3}}$
$\displaystyle \frac{dy}{dx}-\frac{y}{x}=-\frac{e}{x^{4}}y^{-1}$ or the form $\displaystyle \frac{dy}{dx} + P(x)y = Q(x)y^{n}$
This kind of non-linear first order differential equation is a "Bernoulli" equation with n = -1.
Let $\displaystyle v = y^{2}$. Then $\displaystyle \frac{dv}{dx} = 2y \frac{dy}{dx}$, or $\displaystyle \frac{dy}{dx} = \frac{1}{2\sqrt{v}} \frac{dv}{dx}$.
Thus:
$\displaystyle \frac{1}{2\sqrt{v}} \frac{dv}{dx} - \frac{\sqrt{v}}{x} = -\frac{e}{x^4}\frac{1}{\sqrt{v}}$
Thus
$\displaystyle \frac{dv}{dx} - \frac{2}{x}v = -\frac{2e}{x^4}$
which is now a linear first order differential equation, which you can solve using any method you wish. (Remember to back-substitute to get y(x)!)
-Dan
I've tried the polar coordinate substitution you suggested, but I'm not getting any "joy" out of it.