Question

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

• A. 44563
• B. 44593
• C. 44594
• D. 44562

Answer 1

The Correct Option is D

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