Differential Equations Help

[Dash Lambda]

19 minutes ago, djdwosk97 said:

I get #2, but I'm a bit confused on #1. How would I correct it then, how would I solve the diffeq without starter data? 

It gives you the system of DEs, so from there you find the eigenvalues and their associated eigenvectors for the coefficient matrix (also given) and use them to construct the solutions for the system.

For each distinct eigenvalue λ and its associated eigenvector v, [x(t),y(t)]=ve^λt is a solution to the system. As the system is linear, every solution can therefore be expressed as a linear combination of those two solutions; [x(t),y(t)] = C₁v₁e^λ₁t + C₂v₂e^λ₂t

 

The given answer supposed that [x(t),y(t)]=vt is a solution to the system, which is obviously incorrect because that would mean that x(t) and y(t) are linear functions.
What is correct about the given answer is that every 2D vector can be expressed as a linear combination of the eigenvectors, because the eigenvectors are linearly independent and therefore form a basis for R². That's linear algebra, though, so it's not relevant here.

[79wjd]

Okay, last few questions as I'm basically done with the course and I just need to finish up a few things / fix a few things. I'd like to get everything turned in today so that I can take the final tuesday (I can't change anything after the final is submitted). I'm sure you're busy but if you could answer any of these questions today that would be amazing.

 

Question #1

 

 

Section (8.L)


 

 

 

Question #2

 

 

(secition 3.8)

 

Question #3

 

 

 

 

 

Question #4

 

 

 

 

 

Question #5

 

 

 

 

 

Question #6

[79wjd]

Whoops, forgot to @Dash Lambda

[Dash Lambda]

Had to go out for a bit today for mother's day, so I didn't have the time until now -This is the only answer I can give you right now, the others I've gotta spend a little more time on. I'll try to get some more in tomorrow.

 

2: The characteristic equation has a solution of multiplicity 2, so the DE actually has two linearly independent solutions from the same value, which you do by multiplying by t, giving you y=(C₁+C₂t)e^-rt. If what they're referring to is specifically why that one appears -if you get the general solution and differentiate it to get (C₂(1-rt)-C₁r)e^-rt, you see that C₂ must cancel out the C₁ term to achieve the desired initial condition, and must also contain that initial condition, so it becomes an addition of two terms.

[Dash Lambda]

1: a) If both eigenvalues are negative, then the critical point is a sink. This means that all local trajectories are directed towards the point, so f(x,y) increases towards that point.

b) If they're both positive, then the critical point is a source, so all local trajectories point outward.

c) If the eigenvalues are of opposite sign, it is a saddle point (that does give away the answer, but I'm not sure how else I would say it).

 

4: If I get what it's asking: The equilibrium points are the centers and saddle points, where a center indicates a local minimum and a saddle indicates a local maximum. I think that also relates to #3, but either these two questions are poorly worded or I lack context.

 

I haven't done partial Diff Eq yet (actually doing that next semester), so unfortunately I don't know how to answer the last two. I hope the ones I could do were helpful.

[79wjd]

@Dash Lambda

 

Alright, I think that should be all of the diffeq questions I've got forever. Thanks so much. 

[FosterGuy]

This whole thread makes me feel like a supermassive dumbass.

[MrDynamicMan]

On 2017-2-23 at 3:34 PM, Dash Lambda said:

In this instance he actually is directly solving a quadratic equation.

Yeah. However many months down the line we are now I completely understand what he's doing.

[Dash Lambda]

5 hours ago, Kumaresh said:

Are you a college undergrad by any chance? It seems high school D.Es are not as impressive as I thought......

I'm a junior in high school. I took my Diff Eq class at Schoolcraft College, they only had the one -I imagine other colleges could have different levels and types.

 

5 hours ago, Kumaresh said:

@Dash Lambda Plz help me with this problem.

S=1+2+3+4+5+6+7+.........

S=0+1+2+3+4+5+6+.........

Subtracting, we get 1+1+1+1+1+1......=0, how is this possible ? My friend is bugging me, because I just finished explaining how 1+2+3+4+5+.... = -1/12

That's because you're working with a divergent series. Actually, that's how you prove that the series can't be properly manipulated with a stable and linear method of summation.

The -1/12 result is achieved through summation methods which change the series, like Zeta Function Regularization and Ramanujan Summation. If you treat it as a typical summation, it diverges.