Question

Determine order and degree (if defined) of differential equations given in \[y''+2y+siny=0\] .

Answer 1

Hint: At first, we rewrite the differential equation as \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}+siny=0\] . After that, we find the derivative of highest order which is \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]. We note its order. After that, we find the power to which it is raised. It is the degree of the differential equation.

Complete step by step answer:
Order of a differential equation is the order of the highest derivative (also known as differential coefficient) present in the equation. The degree of the differential equation is represented by the power of the highest order derivative in the given differential equation.
The differential equation that we are given in this problem is,
\[y''+2y+siny=0\]
We can rewrite the differential equation as,
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}+siny=0\]
Now, the highest derivative in the above differential equation is \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] . The order of this derivative is nothing but $2$ . We can see that this derivative \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] is raised to the power of nothing but $1$ . So, we can say that the degree of this differential is $1$ .
Thus, we can conclude that the order and degree of the given differential equation are $2$ and $1$ respectively.

Note: We can also solve the problem instead of writing the equation as \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}+siny=0\] . But, for that, we need to know that the order of the derivative of a variable is denoted by the number of apostrophes written to it. Here, for the first term, there are two apostrophes. So, the order is $2$ . For the case of degree, we must be careful. Degree must be that of the highest order derivative and not the overall one. There may be cubes and other higher degrees but that may not be the degree of the differential equation if the degree is not corresponding to the highest order derivative.