Question

Consider the following differential equations.
$D_{1}: y=4 \frac{d y}{d x}+3 x \frac{d x}{d y} ;$
$D_{2}: \frac{d^{2} y}{d x^{2}}=\left(3+\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{4}{3}}$
$D_{3}:\left[1+\left(\frac{d y}{d x}\right)\right]^{2}=\left(\frac{d y}{d x}\right)^{2}$
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,
$D_{1}: y=4 \frac{d y}{d x}+3 x \frac{d x}{d y}$
$\Rightarrow y \frac{d y}{d x}=4\left(\frac{d y}{d x}\right)^{2}+3 x$
$\therefore $ Order $=1$, Degree $=2$
$D_{2}: \frac{d^{2} y}{d x^{2}}=\left(3+\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{4}{3}}$
$\Rightarrow \left(\frac{d^{2} y}{d x^{2}}\right)^{3}=\left(3+\left(\frac{d y}{d x}\right)^{2}\right)^{4}$
$\therefore $ Order $=2$ Degree $=3$
$D_{3}:\left[1+\left(\frac{d y}{d x}\right)\right]^{2}=\left(\frac{d y}{d x}\right)^{2}$
$\Rightarrow 1+ \left(\frac{d y}{d x}\right)^{2}+2 \frac{d y}{d x}=\left(\frac{d y}{d x}\right)^{2}$
$\Rightarrow 1+2 \frac{d y}{d x}=0$
$\therefore $ Order $=1$, Degree $=1$
$\therefore $ Required ratio
$=\frac{1+2+1}{2+3+1}=\frac{4}{6}=\frac{2}{3}$