Q119 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

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 .

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

Hence, the correct answer is option: 'B'.

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