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

The key idea is that order is the highest derivative present, and degree is the power of that highest derivative after the equation is made polynomial in derivatives.

Step 1: Identify the highest derivative. Here it is $y''$ (the second derivative), so the order is 2. …

This is a straightforward problem, but it tests whether you know the definitions cold. Many students get tripped up by the presence of $(y')^2$ and think the degree might be 2. Let’s see why that’s wrong.

The order of a differential equation is simply the highest derivative that appears. Here we have $y''$ (second derivative) and $y'$ (first derivative). The highest is $y''$, so the order is 2.

The degree is trickier. It is defined only when the equation is a polynomial in the derivatives — meaning no fractional powers, no trigonometric functions of derivatives, no absolute values, etc. And even then, the degree is the power of the highest-order derivative after the equation is made polynomial and free of radicals.

Look at the given equation:

It is already a polynomial in $y''$, $y'$, and $y$. The highest derivative $y''$ appears with exponent 1. There is no square root, no fraction, no sine of $y''$ — nothing that would break the polynomial condition. So the degree is simply 1.

Let’s walk through the formal steps:

1. Identify the highest derivative. …

Method: Order and degree when a first derivative is squared

Use this on equations like $y''+(y')^2+2y=0$.

Steps

Step 1: Order = highest derivative present.

$y''$ gives order $2$; the squared $y'$ does not raise it.

Step 2: Confirm polynomial form.

$(y')^2$ is a legitimate polynomial power, and $y''$ carries no radical or transcendental wrapper, so the degree is defined. …

Common Mistakes

Mistake 1: Answering degree $2$ because of $(y')^2$.

Why it's wrong: in $y''+(y')^2+2y=0$ the square is on the lower-order $y'$; the degree is the power of the highest derivative $y''$, which is $1$. Correct approach: order $=2$, degree $=1$. …