Question

The order and degree of the differential equation \({\left( {\frac{{ds}}{{dt}}} \right)^4} + 3s\frac{{{d^2}s}}{{d{t^2}}} = 0\) are______
1. 4, 1
2. 4, 2
3. 2, 1
4. 1, 2
5. None of these

Answer 1

Correct Answer - Option 3 : 2, 1

Order of a differential equation is defined as the order of the highest order derivative and the degree of a differential equation, when it is a polynomial equation in derivatives, is defined as the highest power (positive integral index) of the highest order derivative.

The highest order derivative present in the differential equation is \(\frac{{{d^2}s}}{{d{t^2}}}\), so its order is two. The given differential equation is polynomial equation in its derivatives\(\left( {\frac{{{d^2}s}}{{d{t^2}}}\;\text{and} \ \frac{{ds}}{{dt}}} \right)\). The power raised to \(\frac{{{d^2}s}}{{d{t^2}}}\) is 1, so its degree is one.