# Linear/non-linear

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

Is cosxdy/dx = x non-linear because of cosx?
• Sep 14th 2010, 04:01 AM
CaptainBlack
Quote:

Just a simple question:

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

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.

CB
• Sep 14th 2010, 04:10 AM
Jester
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 AM
CaptainBlack
Quote:

Originally Posted by Danny
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 $\displaystyle \cdots$

I would rather stick with the definition of linearity; less to remember. It is also a property of interest (that solutions satisfy superposition)

CB
• Sep 14th 2010, 05:36 AM
Jester
But aren't you assuming two independent solutions (for a first order ODE?).
• Sep 14th 2010, 05:56 AM
CaptainBlack
Quote:

Originally Posted by Danny
But aren't you assuming two independent solutions (for a first order ODE?).

As it happens, no. The definition still holds. (Alternatively, what else would linear mean?)

CB
• Sep 14th 2010, 06:54 AM
Ackbeet
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 AM
CaptainBlack
Quote:

Originally Posted by Ackbeet
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.

Opps.. that is what I should have said.

CB