Discussion Forum

1)

Consider the following  differential equations.

$D_{1}:y=4\frac{dy}{dx}+3x\frac{dx}{dy};D_{2}:\frac{d^{2}y}{dx^{2}}=\left(3+\left(\frac{dy}{dx}\right)^{2}\right)^{4/3}$

$D_{3}:\left[1+\left(\frac{dy}{dx}\right)\right]^{2}=\left(\frac{dy}{dx}\right)^{2}$

 The ratio of the sum of the orders of $D_{1},D_{2} $ and $D_{3}$  to the sum of their degrees is 

Answer:

Option C

Explanation:

 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

$D_{2}:\frac{d^{2}y}{dx^{2}}=\left(3+\left(\frac{dy}{dx}\right)^{2}\right)^{4/3}$

   =  $\left(\frac{d^{2}y}{dx^{2}}\right)^{3}=\left(3+\left(\frac{dy}{dx}\right)^{2}\right)^{4}$

 $\therefore$    Order=2, Degree=3

$D_{3}:\left[1+\left(\frac{dy}{x}\right)\right]^{2}=\left(\frac{dy}{dx}\right)^{2}$

$\Rightarrow$    $1+\left(\frac{d^{}y}{dx^{}}\right)^{2}+2\frac{dy}{dx}=\left(\frac{dy}{dx}\right)^{2}\Rightarrow1+2 \frac{dy}{dx}=0$ 

 $\therefore$    Order=1, Degree=1

 $\therefore$   Required ratio = $\frac{1+2+1}{2+3+1}=\frac{4}{6}=\frac{2}{3}$