Q.Determine the order and degree, if defined, of the differential equation: $y'' + 2y' + \sin y = 0$

Order = the highest derivative present. Here the highest is $y''$, so the order is $2$.

Degree = the power of the highest-order derivative, provided the equation is a polynomial in all the derivatives. The term $\sin y$ is a function of the dependent variable $y$ — not of a derivative — so it does not spoil that requireme …

What order and degree mean

• Order: the order of the highest derivative that appears.
• Degree: the power to which the highest-order derivative is raised, but only after the equation has been made a polynomial in the derivatives (free of radicals or of transcendental functions of the derivatives).

Apply to $y''+2y'+\sin y=0$

1. The highest derivative present is $y''$, so the order $=2$.
2. Now test for degree. The only unusual term is $\sin y$. The key point: here $\sin$ acts on the dependent variable $y$, not on any derivative such as $\sin y'$ or $\sin y''$. Degree only cares about how the derivatives enter the equation. …

Method: Order and degree when the transcendental function acts on $y$

Use this on equations like $y''+2y'+\sin y=0$, where a $\sin$ appears but wraps the dependent variable.

Steps

Step 1: Order = highest derivative present.

$y''$ gives order $2$.

Step 2: Ask what the transcendental function encloses.

$\sin y$ is a function of $y$ (the dependent variable), not of a derivative. The degree is defined only for a polynomial in the derivatives, and a function of $y$ alone does not violate that. …

Common Mistakes

Mistake 1: Declaring the degree undefined because of $\sin y$.

Why it's wrong: in $y''+2y'+\sin y=0$ the sine wraps the dependent variable $y$, not a derivative, so the polynomial-in-derivatives condition still holds and the degree is defined. Correct approach: degree $=1$. Only $\sin(y')$ or $\sin(y'')$ would make it undefined. …