How do you determine the order of a differential equation? $$t^5y^{(4)} - t^3y'' + 6y = 0$$
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 $ax + c$?
 A: 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
$$t^5 (af(t) + g(t))'''' + t^3(af(t) + g(t))'' +6(af(t) + g(t))\\ = a\big( t^5f^{(4)}(t) +t^3f''(t)+6f(t)\big) + \big(t^5 g^{(4)}(t)+t^3g''(t)+6g(t)\big).$$
A: 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.)
A: 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 $f_4(t)y^{(4)}+f_3(t)y^{(3)}+f_2(t)y^{(2)}+f_1(t)y^{(1)}+f_0(t)y+f(t)=0$
