Order and degree of a differential equation

Question

Here is a question in my book

Find the order and degree of the differential equation $$y=1+\frac{dy}{dx}+\frac{1}{2!}{\left(\frac{dy}{dx}\right)}^2+\frac{1}{3!}{\left(\frac{dy}{dx}\right)}^3+\cdots$$

At first sight we can conclude that the order is $1$ and the degree is undefined as as the power of $\frac{dy}{dx}$ continues to increase and has no limit.However my book gives the following solution

Rewrite the DE as $$y=\exp\left({\frac{dy}{dx}}\right)$$ $$\implies \frac{dy}{dx}=\ln y$$ whose order and degree is 1 .

Now ,I completely agree with this solution however I find it rather counterintuitive to my first line of thought .If the book is correct how can it be justified to prove my intuition was wrong?

Answer 1

The book is indeed incorrect, when you do such modifications on the equation, the order and degree changes. You must read the order and degree of the equation you are given without any simplification. To illustrate what I mean, suppose I give you:

$$ y' =y$$

I could square the above and get:

$$(y')^2 = y^2$$

Now, look at the two equations you may see first equation was first degree and second is second degree.

However, in case of JEE and board exams (Indian syllabus stuff) , usually they want you to simplify before saying degree or order (this is actually wrong though)

Related

Answer 2

Your intuition about the order of given ODE is correct as writing the RHS into closed form function never changes it. But the degree of an ODE is a parameter which is something vague to talk about as one can always disturb it just by squaring both sides.

Though if you constraint yourself to the topic linear ODE (which seems to be true as per your syllabus of JEE) then $y=1+y'+\frac{y'^2}{2!}+....$ is a non-linear ODE which on linearization gives you the linear version : $y'=\log_e y$ has degree $1$ (if you still want to define this parameter).