Question

\[y = px + \sqrt{a^2 p^2 + b^2},\text{ where p} = \frac{dy}{dx}\]

Answer 1

\[y = px + \sqrt{a^2 p^2 + b^2}\]
\[ \Rightarrow y - px = \sqrt{a^2 p^2 + b^2}\]
Squaring both sides, we get
\[ \Rightarrow \left( y - px \right)^2 = a^2 p^2 + b^2 \]
\[ \Rightarrow y^2 - 2pxy + p^2 x^2 = a^2 p^2 + b^2 \]
\[ \Rightarrow \left( x^2 - a^2 \right) p^2 - 2pxy + \left( y^2 - b^2 \right) = 0\]
\[ \Rightarrow \left( x^2 - a^2 \right) \left( \frac{dy}{dx} \right)^2 - 2xy\frac{dy}{dx} + y^2 - b^2 = 0 .............\left[\text{ Substituting p }= \frac{dy}{dx} \right]\]
In this differential equation, the order of the highest order derivative is 1 and its highest power is 2. So, it is a differential equation of order 1 and degree 2.
It is a non-linear differential equation, as its degree is 2, which is greater than 1.