Question:easy

The order and degree of the following differential equation are \( m \) and \( n \), respectively.
\[ \frac{\partial^3 \varphi}{\partial x^3} + \frac{\partial^2 \varphi}{\partial y^2}\frac{\partial \varphi}{\partial x} + \left(\frac{\partial^2 \varphi}{\partial x^2}\right)^2 + \frac{\partial \varphi}{\partial y} = 0 \]
The value of \( (m-n) \) is

Show Hint

Find the highest derivative order first, then check the power of only that highest order term for the degree.
Updated On: Jul 27, 2026
  • \( 2 \)
  • \( 3 \)
  • \( 1 \)
  • \( 0 \)
Show Solution

The Correct Option is A

Solution and Explanation

Order and degree both describe a differential equation, but they measure different things. Order counts how many times the function has been differentiated in the highest term present; degree counts the power on that highest order term only, after clearing any roots or fractions involving derivatives.

Scanning the given equation, the term $\partial^3\varphi/\partial x^3$ carries the most derivatives, three of them, so the order is 3. Every other term, including the squared term $(\partial^2\varphi/\partial x^2)^2$, involves fewer derivatives (order 2), so that squaring does not enter the degree calculation at all, it only matters for a term sitting at the highest order.

Since the order 3 term shows up to the first power only, the degree is 1. That makes $m = 3$ and $n = 1$.

  1. 2: matches $m - n = 3 - 1 = 2$, so this is correct.
  2. 3: would be correct only if the degree were 0, which is not the case here.
  3. 1: would follow if the order were 2, but the third order term rules that out.
  4. 0: would mean order and degree are equal, which they are not.

So $m - n = 2$, option (A).

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