Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$t^5y^{(4)} - t^3y'' + 6y = 0$$

the answer is fourth order, but I don't understand why exactly is it because of $y^{(4)}$? If so, is $y^{(4)}$ equivalent to $y''''$?

also, it says the equation is linear, but how is that possible if the exponent of $t$ is $5$ (and not 1)? Shouldn't a linear equation be of the form $ax + c$?

share|cite|improve this question
ah its '' my bad. y'' – Gladstone Asder May 2 '13 at 2:02
A differential equation is linear if it is of the form $$a_n(t)y^{(n)} + a_{n-1}(t)y^{(n-1)} + \cdots + a_1(t)y' + a_0(t) y = f(t)$$ where $a_i$ and $f(t)$ are all functions of $t$. – Kris Williams May 2 '13 at 2:09
ah nice thanks. – Gladstone Asder May 2 '13 at 2:11
isn't ai a constant and not a function of t? – Gladstone Asder May 2 '13 at 2:14
@GladstoneAsder The definition of a linear DE allows for the $a_i$ to be functions of $t$. – apnorton May 2 '13 at 2:18
up vote 2 down vote accepted

Yes, $y'''' = y^{(4)}$. So it's fourth order because the highest order derivative is $4$.

Linear functions have the form $ax + c$. This is a linear operator: it doesn't eat numbers, it eats functions. So to see it's linear, take two different functions, $f$ and $g$, and a constant $a$, and verify that $$t^5 (af(t) + g(t))'''' + t^3(af(t) + g(t))'' +6(af(t) + g(t))\\ = a\big( t^5f^{(4)}(t) +t^3f''(t)+6f(t)\big) + \big(t^5 g^{(4)}(t)+t^3g''(t)+6g(t)\big).$$

share|cite|improve this answer
another question can you transform the equation into y = ax + c form? – Gladstone Asder May 2 '13 at 2:06
Instead of thinking of a line as $y = ax + c$, think of it as a function $f(x) = ax + c$. This is a linear equation because, for any numbers $b, x, y$, $f(bx + y) = bf(x) + f(y)$. That's how you should think of this differential equation. – Neal May 2 '13 at 2:17

You're on the right track. Basically, $y^{(4)}$ is shorthand for $y''''$. This notation arises because counting little tickmarks is bothersome, to say the least. For example, imagine if I wanted $\frac{d^{104}y}{dx^{104}}$. You'd have to draw $104$ little tickmarks, or you could simply say $y^{(104)}$.

And in general, the order of a differential equation is the order of the greatest derivative in the problem. So, $y^{(4)} = y$ is a fourth-order, and $y' = y$ is a first order.

Regarding linear differential equations: In determining order of differential equations, we don't care about what $t$ does. The same is true for determining if a DE is linear or not. We want a linear combination of the derivatives of the function. So, the first two examples below are linear, the second two are not: $$y' + y'' + y = t$$ $$t^4y'' + \cos t + y = 1$$ $$\cos(y'') + y = 3$$ $$(y')^2 + y = t$$ (Note that for the last one, I'm denoting exponentiation, not the order of the derivative. The difference is the lack of parenthesis in the superscript.)

share|cite|improve this answer

Its fourth order because the maximum derivative is the fourth, and $y^{(4)}$ is the same as $y''''$.

Its linear because you can write as $f_4(t)y^{(4)}+f_3(t)y^{(3)}+f_2(t)y^{(2)}+f_1(t)y^{(1)}+f_0(t)y+f(t)=0$

share|cite|improve this answer
so y'''' = y^(4)? – Gladstone Asder May 2 '13 at 2:05
and why is it linear? – Gladstone Asder May 2 '13 at 2:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.