The order and the degree of the differential equation $y=p x+\sqrt{a^2 p^2+b^2}$, ( where $p=\frac{d y}{d…
The order and the degree of the differential equation $y=p x+\sqrt{a^2 p^2+b^2}$, ( where $p=\frac{d y}{d x}$ ) are respectively .
2,1
1,2
1,1
2,2
Solution
Given differential equation,
$
\begin{aligned}
y & =p x+\sqrt{a^2 p^2+b^2}, \quad\left[\text { where, } p=\frac{d y}{d x}\right] \\
\Rightarrow \sqrt{a^2 p^2+b^2} & =(y-p x) \\
\Rightarrow \quad a^2 p^2+b^2 & =y^2+p^2 x^2-2 x y p \\
\Rightarrow\left(x^2-a^2\right) p^2 & -2 x y p+y^2-b^2=0
\end{aligned}
$
So, order and degree of the given differential equation is 1 and 2 respectively