Question

The differential equation whose solution is $A{x}^2+B{y}^2=1$, where A and B are arbitrary constants is of

• A. second order and second degree
• B. first order and second degree
• C. first order and first degree
• D. second order and first degree

Answer 1

The correct option is D second order and first degree

$A{x}^2+B{y}^2=1$ ...... $\left(i\right)$
Differentiating $\left(i\right)$ w.r.t $x,$ we get
$2xA+2By\frac{dy}{dx}=0$ $\Rightarrow$ $xA+By\frac{dy}{dx}=0$ ........ $\left(ii\right)$
Differentiating $\left(ii\right)$ again w.r.t. $x$ we get
$A+By\frac{d^{2}y}{d{x}^2}+B{\left(\frac{dy}{dx}\right)}^2=0$ ........ $\left(iii\right)$
From $\left(ii\right)$ and $\left(iii\right)$, we get
$x[-By\frac{d^{2}y}{d{x}^2}-B{\left(\frac{dy}{dx}\right)}^2]+By\frac{dy}{dx}=0$
$\therefore$ Order $=2$ and degree $=1$.