I'm too embarrassed to ask the TA this. I feel like I've forgotten something fundamental in math

I'm too embarrassed to ask the TA this. I feel like I've forgotten something fundamental in math.

The second one. It's homogenous either way since it includes your variable y (and it's derivative) only once.

You can do it either way.

Easiest is this:

dy/2y = 1/5 dx

1/2 ln y = 1/5x + c

y = Ce^(2x/5)

Am i retarded? I don't see how you algebra'd your way to option A. I only understand option B.

Linear,
Solved by all Ce^{2x/5} for C in R

I am retarded. They should be equal.

He's saying 2y/2y = 1

Then subtracting from both sides

multiply both sides of the equation by (2y)^-1

I'm not sure what your question is. Are you asking if we can classify an equation as linear if it can be transformed to a non-linear equation? Well, yes. The term "linear" describes the form of equation, it's irrelevant if some non-linear equation has the same solutions. A simple example:
[math]x-1=0[/math]
[math]x^3-1=0[/math]
The first equation is obviously linear, despite the fact that the second equation has the same real number solution.

they are all fucking linear because they are all the same fucking ODE.

go google "linear ODE" or read your book.

This

Long story short:
An ODE is linear if it can be represented with f(x) in front of the dependent variable (y) or its derivatives.
Of course, you could turn a linear ODE into a non-linear one. But, why? As you will see in 2 weeks max, non-linear ones are almost impossible to solve unless in very rare cases.

fucking brainlets these days ruining Veeky Forums

Out, now.

>too embarrassed to ask the TA
...therefore too delicate for college.
Drop out, join a monastery.

You're x's are terrible. Get better handwriting, faggot.

No one brought this up, but A and B are not exactly the same ODE. Be careful, B has the solution y=0, but A does not.

Wtf the fuck am I reading

Ya had a feeling this was the answer after I posted this.

Just because an equation can be written as non-linear doesn't mean it is.

It's obvious now, but for some reason it was confusing me.

You have de L[y]=0
linear iff
L[ay+bz]=aL[y]+bL[z]

How the bloody hell did you get option 1? This is like grade 7 stuff

>dy/dx dependent on anything
>lel is it linear

brainlet identified

>not just separating the DE