The order and degree of the following differential equation :
\( y=px+\sqrt{a^2p^2+b^2} \), where \( p=\displaystyle\frac{dy}{dx} \), are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Order = 1 : degree = 2
Only the first derivative \( p=\frac{dy}{dx} \) appears, so the order is 1.
The degree is read only after the square root is removed:
\( y-px=\sqrt{a^2p^2+b^2} \)
\( (y-px)^2=a^2p^2+b^2 \)
\( y^2-2xyp+x^2p^2=a^2p^2+b^2 \)
\( (x^2-a^2)p^2-2xyp+y^2-b^2=0 \)
This is a polynomial in p whose highest power is 2, so the degree is 2.
Why the other options are incorrect
Option (1) Order = 2 : degree = 2 — Incorrect. There is no second derivative in the equation, so the order cannot be 2.
Option (2) Order = 1 : degree = 1 — Incorrect. The degree 1 is read before removing the square root; after squaring, p appears with power 2.
Option (4) Order = 2 : degree = 1 — Incorrect. The order is 1 (only dy/dx appears), and the degree after removing the root is 2.