Question

Form the differential equation corresponding to $a{x}^2+b{y}^2=1,(a,b)$.

Answer 1

$\begin{array}{c}(a,b)twoarbitrearycons\tan t\\ Thedifferentiatetwotimesw.r.tox\\ a{x}^2=1-b{y}^2\\ a^2=\frac{1}{x^2}+\frac{b{y}^2}{x^2}\\ \frac{d{a}^2}{dx}=\frac{d{x}^2}{dx}+b\left[y^2\frac{d{x}^{-2}}{dx}+x^{-2}\frac{d{y}^2}{dx}\right]\\ 0=-2x^{-3}+b\left[\frac{-2y^2}{x^3}+2y^{-2}dy/dx\right]\\ 2x^{-3}=b\left[\frac{2y^2}{x^3}+\frac{2y}{x^2}dy/dx\right]\\ \frac{2}{x^3}=\frac{2b}{x^2}\left[\frac{y^2}{x}+ydy/dx\right]\\ \frac{1}{x}=b\left[ydy/dx-\frac{y^2}{x}\right]\\ \frac{1}{x\left(ydy/dx-\frac{y^{2}y}{x}=b\right)}\\ \frac{1}{\begin{array}{c}x\left(xydy/dx-y^2\right)\\ \frac{1}{b}=\left(xydy/dx-y^2\right)\end{array}}=b\\ \frac{d-\left(\frac{1}{b}\right)}{dx}=\frac{\left(xy/dx-y^2\right)}{dx}\\ 0=\left[xy\frac{d\left(dy/dx\right)}{dx}+dy/dx.\frac{d\left(xy\right)}{dx}-=\frac{d{y}^2}{dx}\right]\\ 0=xy\frac{d^{2}y}{d{x}^2}+dy/dx\left[xdy/dx+y\frac{dx}{dx}\right]-\left[2ydy/dx\right]\\ \Rightarrow 0=xy\frac{d{x}^2}{d{x}^2}+x{\left(dy/dx\right)}^2+ydy/dx-2ydy/dx\\ \Rightarrow xy\frac{d^{2}y}{d{x}^2}+dy/dx\left[xdy/dx-y\right]=0\end{array}$