Identifying the Degree of a Differential Equation

[KD1729]

How Can we define the degree of differential equation ?
What is the degree of \(\displaystyle \left(\frac{{d}^{2}y}{d{x}^{2}}\right)^{\!{2}}\left(\frac{dy}{dx}\right)^{\!{3}} +\left(\frac{dy}{dx}\right)^{\!{1}} =0\) ??(Wondering)

 

[MarkFL]

The degree of a differential equation is the power of the highest order derivative in the equation. Based on this what would you say the degree of the given ODE is?

 

[KD1729]

The degree of a differential equation is the power of the highest order derivative in the equation. Based on this what would you say the degree of the given ODE is?

Then it Should be one.

But, My maths Teacher said it is equal to the 2+3=5.
He said that for any term degree is the sum of exponents of its variables & degree of a equation is the highest degree term. Which is right ?(Sadface)

 

[MarkFL]

I would look at the derivative in red:

\(\displaystyle {\color{red}\left(\frac{{d}^{2}y}{d{x}^{2}}\right)^{\!{2}}}\left(\frac{dy}{dx}\right)^{\!{3}} +\left(\frac{dy}{dx}\right)^{\!{1}} =0\)

This is the highest order derivative present in the equation and the degree of this derivative is two, thus I would say the degree of the ODE is two.

 

[blue_raver22]

The degree of a differential equation is determined by the highest power of the highest order derivative present in the equation. In this case, the highest order derivative is \frac{{d}^{2}y}{d{x}^{2}} and it is raised to the power of 2. Therefore, the degree of this differential equation is 2.