Just a simple question:

Is cosxdy/dx = x non-linear because of cosx?

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- Sep 14th 2010, 03:30 AMCookieCLinear/non-linear
Just a simple question:

Is cosxdy/dx = x non-linear because of cosx? - Sep 14th 2010, 04:01 AMCaptainBlack
An differential operator $\displaystyle $$ L$ is linear iff for any two solutions $\displaystyle y_1(x)$ and $\displaystyle y_2(x)$ of $\displaystyle Ly=0$ and any constants $\displaystyle \alpha, \beta \in \mathbb{C}$ then $\displaystyle y(x)=\alpha y_1(x)+\beta y_2(x)$ is also a solution. An ODE is linear if it is of the form $\displaystyle Ly=f(x),\ L$ a linear differential operator

Now you can check for yourself.

(the answer is: it's linear)

CB - Sep 14th 2010, 04:10 AMJester
Any first ODE that can be written as

$\displaystyle \dfrac{dy}{dx} = P(x) y + Q(x)$

i.e. linear in both $\displaystyle y$ and $\displaystyle y'$ is a linear ODE. Your ODE is

$\displaystyle \dfrac{dy}{dx} = \sec x \, y$ so it's $\displaystyle \cdots$ - Sep 14th 2010, 05:23 AMCaptainBlack
- Sep 14th 2010, 05:36 AMJester
But aren't you assuming two independent solutions (for a first order ODE?).

- Sep 14th 2010, 05:56 AMCaptainBlack
- Sep 14th 2010, 06:54 AMAckbeet
According to the wiki, a linear differential equation is of the form

$\displaystyle Ly(x)=f(x),$ where $\displaystyle L$ is a linear differential operator.

That is, you apply CB's definition of linearity to the*operator*, not necessarily to the entire equation. If the equation is homogeneous, they'll turn out to be the same thing. But even a non-homogeneous linear DE does not obey the superposition principle except in its homogeneous solution.

This is my understanding. - Sep 14th 2010, 06:57 AMCaptainBlack