Discussion
The order and degree of the following differential equation are \(m\) and \(n\), respectively.
\[ \begin{aligned} &\frac{\partial^{3}\varphi}{\partial x^{3}} + \frac{\partial^{2}\varphi}{\partial y^{2}}\frac{\partial\varphi}{\partial x} \\ &+ \left(\frac{\partial^{2}\varphi}{\partial x^{2}}\right)^{2} + \frac{\partial\varphi}{\partial y} = 0 \end{aligned} \]
The value of \((m - n)\) is
Answer: A
The value of \(m - n\) is 2.
The order is the highest derivative present, \(\partial^{3}\varphi/\partial x^{3}\), so \(m = 3\). The degree is the power of that highest-order derivative once the equation is polynomial in its derivatives; \(\partial^{3}\varphi/\partial x^{3}\) appears to the first power, so \(n = 1\).
Hence \(m - n = 2\). The square on \(\partial^{2}\varphi/\partial x^{2}\) is a trap: it is not the highest-order term, so it does not set the degree. Taking the degree as 2 gives 1, and taking the order as 2 gives 0 or 1.
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