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

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 $ax + c$?

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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 $$t^5 (af(t) + g(t))'''' + t^3(af(t) + g(t))'' +6(af(t) + g(t))\\ = a\big( t^5f^{(4)}(t) +t^3f''(t)+6f(t)\big) + \big(t^5 g^{(4)}(t)+t^3g''(t)+6g(t)\big).$$

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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.)

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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 $f_4(t)y^{(4)}+f_3(t)y^{(3)}+f_2(t)y^{(2)}+f_1(t)y^{(1)}+f_0(t)y+f(t)=0$

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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

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