I'm too embarrassed to ask the TA this. I feel like I've

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

>>8631143
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)

>>8631143
Am i retarded? I don't see how you algebra'd your way to option A. I only understand option B.

>>8631143
Linear,
Solved by all Ce^{2x/5} for C in R

>>8631272
I am retarded. They should be equal.

>>8631272
He's saying 2y/2y = 1

Then subtracting from both sides

>>8631272
multiply both sides of the equation by (2y)^-1

>>8631143
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.

>>8631143
they are all fucking linear because they are all the same fucking ODE.

go google "linear ODE" or read your book.

>>8631433
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.

File: WechatIM12312312313312G1.jpg (43KB, 960x1280px) Image search: [Google]

43KB, 960x1280px

fucking brainlets these days ruining /sci/

>>8632829
Out, now.

>>8631143
>too embarrassed to ask the TA
...therefore too delicate for college.
Drop out, join a monastery.

>>8632829
You're x's are terrible. Get better handwriting, faggot.

>>8631143
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.

>>8632829
Wtf the fuck am I reading

>>8631742
>>8631361
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]

>>8631143
How the bloody hell did you get option 1? This is like grade 7 stuff

>>8631143
>dy/dx dependent on anything
>lel is it linear

brainlet identified

>not just separating the DE