# How to know if the equation is linear?

According to my maths book an equations is linear if,

there are no products of the function and neither the function or its derivatives occur to any power other than the first power.

It should be in form of $$a_n(t)y^{(n)}(t)+a_{n-1}(t)y^{(n-1)}(t)+\cdots+a_1(t)y'(t)+a_0(t)y(t)=g(t)$$

I do understand the first power thing i.e for e.g., shouldn't be like this $(dy/dx)^2$ but can't really understand the other product thing. Please help. Thanks.

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## 3 Answers

There's no $(dy/dx)^2$ (first derivative squared) in your canonical form. There is a $d^2y/dx^2$ (second derivative) hidden at the end of the "$\cdots$", but that is a different thing.

A linear ODE can have terms is a constant times any higher derivative, but there can only be one derivative in each term -- that is, there cannot be two or more (different or same) derivatives multiplied by each other.

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Here is a "positive" version of the criterion:

An ODE is linear if given any two solutions $y_1(\cdot)$, $\ y_2(\cdot)$ the sum $\lambda y_1(\cdot)+\mu y_2(\cdot)$ is again a solution whenever $\lambda+\mu=1$. This test can easily be performed by looking at the equation.

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Like this point of view. – Babak S. Dec 28 '12 at 10:55
Yes, almost -- except that the OP's definition allows non-homogeneous equations (the $g(t)$ on the right). – Henning Makholm Dec 28 '12 at 11:16
@Henning Makholm: Exactly because of that my answer sounds so sophisticated $\ldots$ – Christian Blatter Dec 28 '12 at 11:25
Oh, I missed the $\lambda+\mu=1$ condition. Right, then. – Henning Makholm Dec 28 '12 at 11:28

Sometimes in Mathematics one uses the same name for two different things or two different definitions which are not related whatsoever. But here this is not the case. I am sure you know what linear means in terms of maps between vector spaces. So if $V$ is a vector space and $A:V\rightarrow V$ is a linear map, then it satisfies $A(\alpha \mathbf v + \beta \mathbf w) = \alpha A(\mathbf v) + \beta A(\mathbf w)$.

Now something like $d/dx$ can be viewed as a map on a vector space. There are actually many different vector spaces that would work, but I suggest you just think of the vector space given by all smooth functions from $\mathbb R \rightarrow \mathbb R$. This is a vector space over $\mathbb R$ and $dy/dx$ is a linear map on it. But an expression as $(d/dx)^2$ is not linear on this vector space. Recall that $(dy/dx)^2$ acts as follows:

$$(d/dx)^2 (f) = (d/dx)(f) (d/dx)(f) = f' f'$$

Note that something like $a_n(t) d^n/dx^n + \dots + a_1(t) d/dx + a_0(t)$ is also a linear map on this vector space (if al the a_i(t) are smooth, otherwise you need to change the vector space a bit). So one way to figure out wether a ODE is linear or not is to try to find out if it acts linear on smooth functions.

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