Question

The order and degree of the differential equation y =$\frac{dp}{dx}x+\sqrt{a^2{p}^2+b^2}$ where $p=\frac{dy}{dx}$(here $a$ and $b$ are arbitrary constants) respectively are

• A. 1, 2
• B. 2, 1
• C. 2, 2
• D. 1, 1

Answer 1

Answer: (C)

Given differential equation is
$y=\left(\frac{dp}{dx}\right)x+\sqrt{a^2{p}^2+b^2},\left(p=\frac{dy}{dx}\right)$
$\Rightarrow y=\left[\frac{d}{dx}\left(\frac{dy}{dx}\right)\right]\cdot x+\sqrt{a^2{\left(\frac{dy}{dx}\right)}^2+b^2}$
$\Rightarrow \left(y-x\frac{d^{2}y}{d{x}^2}\right)=\sqrt{a^2{\left(\frac{dy}{dx}\right)}^2+b^2}$
$\Rightarrow {\left(y-x\frac{d^{2}y}{d{x}^2}\right)}^2=a^2{\left(\frac{dy}{dx}\right)}^2+b^2$
Hence, order $=2$ and degree $=2$