Question

The differential equation satisfied by ax² + by² = 1 is
(a) xyy₂ + y₁² + yy₁ = 0
(b) xyy₂ + xy₁² − yy₁ = 0
(c) xyy₂ − xy₁² + yy₁ = 0
(d) none of these

Answer 1

(b) xyy₂ + xy₁² − yy₁ = 0

We have,
ax² + by² = 1 .....(1)
Differentiating both sides of (1) with respect to x, we get
$2ax+2by\frac{dy}{dx}=0.....\left(2\right)$
Differentiating both sides of (2) with respect to x, we get
$$\begin{gathered} 2a+2b{\left(\frac{dy}{dx}\right)}^2+2by\frac{d^{2}y}{d{x}^2}=0 \\ \Rightarrow 2b\left[y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2\right]=-2a \\ \Rightarrow \left[y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2\right]=-\frac{2a}{2b} \\ \Rightarrow \left[y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2\right]=-\left(-\frac{y}{x}\frac{dy}{dx}\right)\left[\text{Using}\left(2\right)\right] \\ \Rightarrow x\left[y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2\right]=y\frac{dy}{dx} \\ \Rightarrow xy\frac{d^{2}y}{d{x}^2}+x{\left(\frac{dy}{dx}\right)}^2=y\frac{dy}{dx} \\ \Rightarrow xy\frac{d^{2}y}{d{x}^2}+x{\left(\frac{dy}{dx}\right)}^2-y\frac{dy}{dx}=0 \\ \Rightarrow xy{y}_2+x{\left(y_1\right)}^2-y{y}_1=0 \end{gathered}$$