Question

Consider the following differential equations.
$D_1:y=4\frac{dy}{dx}+3x\frac{dx}{dy};$
$D_2:\frac{d^{2}y}{d{x}^2}={\left(3+{\left(\frac{dy}{dx}\right)}^2\right)}^{\frac{4}{3}}$
$D_3:{\left[1+\left(\frac{dy}{dx}\right)\right]}^2={\left(\frac{dy}{dx}\right)}^2$
The ratio of the sum of the orders of $D_1,D_2$ and $D_3$ to the sum of their degrees is

• A. 1: 2
• B. 1: 1
• C. 2: 3
• D. 3: 2

Answer 1

Answer: (C)

We have,
$D_1:y=4\frac{dy}{dx}+3x\frac{dx}{dy}$
$\Rightarrow y\frac{dy}{dx}=4{\left(\frac{dy}{dx}\right)}^2+3x$
$\therefore$ Order $=1$, Degree $=2$
$D_2:\frac{d^{2}y}{d{x}^2}={\left(3+{\left(\frac{dy}{dx}\right)}^2\right)}^{\frac{4}{3}}$
$\Rightarrow {\left(\frac{d^{2}y}{d{x}^2}\right)}^3={\left(3+{\left(\frac{dy}{dx}\right)}^2\right)}^4$
$\therefore$ Order $=2$ Degree $=3$
$D_3:{\left[1+\left(\frac{dy}{dx}\right)\right]}^2={\left(\frac{dy}{dx}\right)}^2$
$\Rightarrow 1+{\left(\frac{dy}{dx}\right)}^2+2\frac{dy}{dx}={\left(\frac{dy}{dx}\right)}^2$
$\Rightarrow 1+2\frac{dy}{dx}=0$
$\therefore$ Order $=1$, Degree $=1$
$\therefore$ Required ratio
$=\frac{1+2+1}{2+3+1}=\frac{4}{6}=\frac{2}{3}$