Differential Equations

[Necronic]

So.....I've been somewhat unhappy with my career and have decided to go back for an advanced degree in engineering to expand my options. That means I have to take some classes. Right now I am taking Engineering Mathematics, which, it turns out, is Differential Equations. I have avoided this math for years as well as I could, but now....now I have to learn it!

My first quiz (due in a week) throws me right into the monstrosity of it. What kills me the most is that this dude uses a pretty slanged out Lagrange symbology, which I don't particularly like. Well...actually what kills me the most is that its been 10 years sincce I have taken calculus. What was I thinking?

Anyways, that's not my question. Here's my question.

Identify the differential equation solved by

y=e^(-6x)

A) y'' + 10y' - 4y = 0
B) y'' - 10y' + 24 y = 0
C) y'' + 6y' + 24y = 0
D) y'' - 10 y' - 24y = 0
E) y'' + 10y' + 24y = 0
F) None of the above

In liebniz notation I am pretty sure the first one is

d^2y/dx^@ + 10dy/dx - 4y = 0

but I can't remember how to integrate something like this. I know you can integrate by parts, but what's the integral of y?

 

[tegid]

Oh god why did I do this?

Hey, you don't need to integrate it, you have the solution!

You need to calculate the first and second derivatives of the exponential and input them in the different options. It should only satisfy one of the equations.

About integrating it, you can't integrate y because you don't know what it is. Usually the general solution to these equations is a combination of exponentials or polynomials, so you just put y=e^(a*x) or y=x^n and find a or n.

 

[Piotyr]

Oh god why did I do this?

You don't have to worry about integration in this one, just find the first and second derivatives. Then plug all 3 into each equation and figure out which (if any) is satisfied.

 

[Necronic]

Oh god why did I do this?

riggggghhhhhhht. I was working totally backwards. I just did the derivatives of the solution and got it. Answer is .... E I think, its either that one or D. Don't have the work in front of me, but the signs work on one of those.

 

[evilmike]

Oh god why did I do this?

riggggghhhhhhht. I was working totally backwards. I just did the derivatives of the solution and got it. Answer is .... E I think, its either that one or D. Don't have the work in front of me, but the signs work on one of those.

If I had to guess, I'd say E. (Given that it's late and I'm not checking my work on paper.)

I remember that Diff EQs not being what I expected at all. It felt much more cookbook -- the derivatives are related like A, therefore it is likely that it is a equation of the form B, and then apply the boundary values to get the specific equation.

 

[Math242]

Oh god why did I do this?

head asplode.

 

[LordRendar]

Oh god why did I do this?

Oh hai! I came for the drama.
Omg,whats this? Numbers? Equations?



head asplode.

 

[Adam]

Oh god why did I do this?

I am in Marketing, ergo I believe the correct answer is "Chartreuse"

 

[Ross]

Oh god why did I do this?

y = e^-6x
y' = -6e^-6x
y'' = 36e^-6x

E) y'' + 10y' + 24y = 0

(36e^-6x) + 10(-6e^-6x) + 24(e^-6x) = 0

[36 + 10(-6) +24](e^-6x) = 0

(36 - 60 + 24)(e^-6x) = 0

(0)(e^-6x) = 0

0 = 0 //

The easiest way to approach this type of question is to 1) write out the derivatives, then 2) quickly calculate the coefficients to determine plausible answers. This question was pretty straight-forward, so applying those two steps worked out fine. In more complicated Diff Eq questions, this will at least get you started towards a possible solution.

(edited for explanation)

 

[IronBrig4]

Oh god why did I do this?

This makes me so glad I'm a history grad student. Math hard. Me no like.

 

[Seraphyn]

Oh god why did I do this?

This is the part of math I always failed at. Granted a big reason was that I didn't study, but man, math = head asplode.

 

[Necronic]

Oh god why did I do this?

Now that I've done a couple of them they are pretty easy. The ones that are messing me up now are the limits ones. Like this:

Suppose y=y(x) is the solution to the initial value problem

[y'-2y=4, y(0) = 4]

compute

lim y(x)
x-> inf

 

[tegid]

Oh god why did I do this?

Do you know how to solve that equation? It's not too hard and if you are able to solve it then the solution to the limit is very easy, but I think you are supposed to do it without finding y(x)...

 

[evilmike]

Oh god why did I do this?

Do you know how to solve that equation? It's not too hard and if you are able to solve it then the solution to the limit is very easy, but I think you are supposed to do it without finding y(x)...

I imagine that solution for the equation can be found from the information provided. It's just a matter of finding a general solution based on the relationship between the derivatives and then applying the boundary value to it.

Necronic, is finding y(x) the problem, or the limit?

 

[tegid]

Oh god why did I do this?

I know it can be done, it's actually pretty easy. But it's weird because it's such a jump in difficulty from the question in the OP. Also, the limit is trivial if you explicitly know y(x).

 

[evilmike]

Oh god why did I do this?

I know it can be done, it's actually pretty easy. But it's weird because it's such a jump in difficulty from the question in the OP. Also, the limit is trivial if you explicitly know y(x).

I'm just not sure how you'd get the answer without finding y(x).

 

[tegid]

Oh god why did I do this?

Me neither.

So, how to solve it:

in equations of the type

diff eq=f(x), you can find a solution which is a linear combination of two solutions: the homogeneous solution, which is the solution of diff eq=0 (as in the OP), plus a particular solution found ad hoc so that the equality with f(x) is satisfied.

In this case, the homogeneous solution to y'-2y=0 is an exponential, y=e^Ax. to find A, we put the function into the equation: (A-2)*e^Ax=0 => A=2.=> y_h=exp(2*x)
The particular solution is very easy to find: since f(x) is a constant, it can just be a constant B the derivative of which is zero: 0-2B=4 => B=-2.

The general solution is, then, y(x)=C exp(2*x) - 2. To find C you only need to apply the initial condition. But you don't need to, you can already compute the limit from here (which makes me wonder if this is the correct answer or there is an easier one).

 

[evilmike]

Oh god why did I do this?

Me neither.

So, how to solve it:

in equations of the type

diff eq=f(x), you can find a solution which is a linear combination of two solutions: the homogeneous solution, which is the solution of diff eq=0 (as in the OP), plus a particular solution found ad hoc so that the equality with f(x) is satisfied.

In this case, the homogeneous solution to y'-2y=0 is an exponential, y=e^Ax. to find A, we put the function into the equation: (A-2)*e^Ax=0 => A=2.=> y_h=exp(2*x)
The particular solution is very easy to find: since f(x) is a constant, it can just be a constant B the derivative of which is zero: 0-2B=4 => B=-2.

The general solution is, then, y(x)=C exp(2*x) - 2. To find C you only need to apply the initial condition. But you don't need to, you can already compute the limit from here (which makes me wonder if this is the correct answer or there is an easier one).

I think that's exactly what needs to be done. Technically, you still need to find C if they make a distinction between +∞ and -∞.

 

[tegid]

Oh god why did I do this?

Ah, yes, you are right!

Necronic, let us know if you need any more help!

 

[Necronic]

Oh god why did I do this?

ahhhhh ok. I think I get it. I mean, I definitely get the exponential solution, I guess as you do this more you start to recognize certain forms. What I don't get is how you set y'-2y=0 like that. Was that just so you could solve the exponential, then add the -2 constant to get the 4? Maths are hard.

edit: just to check

y'=2Ce^2x
y=Ce^2x - 2

y(0) = 4 = Ce^0-2 = C - 2
C = 6?

 

[tegid]

Oh god why did I do this?

Yes!

If I'm right, whenever you have an ordinary differential equation with a term without an y, you can forget that term, find a solution and then find an ad hoc solution for that f(x) term. The general solution will be a combination of both like what you just did.

Maybe this only works if you don't have higher orders in y, i.e. y^2, y*y' or things like that. I'm not sure. (For simple equations like the ones you've shown us so far, it definitely works).

 

[evilmike]

Oh god why did I do this?

What I don't get is how you set y'-2y=0 like that.

That type of equation, the relationship between derivatives, is part of the given for a differential equation problem. When you are analyzing a problem, sometimes it is not obvious what the actual function is, but a relationship between the function and one or more of its derivatives will be apparent.

For example, in radioactive decay, the rate of decay of a sample is proportional to the mass of that sample.

So, for the mass function, M(t): dM/dt = k * M(t) with M(0) = A

By applying differential equations, we know that M(t) must be of the form A*e^(k*t). We can use the M(0) boundary condition M(0) to find A, but we will need another piece of data to solve for k.

 

[tegid]

Oh god why did I do this?

I think he means setting the right hand side of equation to zero. Still, good explanation

 

[evilmike]

Oh god why did I do this?

Ah, ok. I forgot that the original equation was y'-2y=4. When I worked through it, I found a slightly different equation as a starting point and did not set it to zero.

And, thanks. It's something I keep coming back to because it's what helped me get through Diff EQs -- realizing that there was a step in solving differential equations that was unlike what I had run into in Algebra and Calculus.

 

[fade]

Oh god why did I do this?

Pbbt. If you're an engineer, the proper way to approach it would be to completely eschew the analytical solution, and expand the unknown derivatives over a twice differentiable space. Since that's unsolvable, too, expand that space even more generally to orthonormal linear function space, and define an orthogonal space of equal type and dimension (why not, says Galerkin). Piece out your solution domain into millions of small bits and find the linear coefficients that define locally linear approximations of the true solution over each little bit. It'll cost you an hour of computer time (just to run the solution) and several gigs of RAM, but it'll get the job done. AND it's reusable for any problem. That's an engineering win!

 

[Necronic]

Oh god why did I do this?

@ Fade- yep. You can spend hours trying to get the analytical solution for some differential equation, or just put it into the appropriate form, jump into excel (or some higher dimensional program like matlab depending), set your dx or dt or whatever to small, and run it. Sadly though, that's not a correct answer for my diff eq prof. Numerical methods for the win.

Anyways, hit a problem I am having some trouble with, thought yall could help again. I'm getting a lot better at this, but I am still rusty on my integral calc. Right now I am messing with linear diff eqs. The general form is :

y' + p(x) y = q(x)


here's the problem

y' + y * cot x = csc^2 x
p(x) = cot x
h(x) = ln(sin x)
e^h(x) = sin x

Multiply the starting equation by e^h(x) and get

sin x[y' + y cot x] = sin x * csc^2 x
[y*sin x]' = csc x = sin ^-1 x

now this is where I am wondering if I have done it right, because the integral for that is not simple:

y*sin x = integral(csc x) = - ln |csc x + cot x| + C

y = sin x * -ln(csc x + cot x) + C sin x

There has to be a more simple form of that. Any thoughts?

 

[Necronic]

Oh god why did I do this?

I'm going to keep updating this as I go, and to be honest, the title hasn't changed. I turned in the first homework assignment last night. I had 20 pages and spent almost 10 hours on it and I think I still only got like 1/2 of them. There were almost 30 questions, and most of them took almost a page. That was brutal. I am gaining a whole new level of respect for engineers and mathmeticians.

In some ways the class is horrendous, but in others I am starting to get it. I understand 1st order linears, bernouli, and seperation of variables. Orthogonal sets aren't hard either, nor are the other applications. Ultimately I get how to do the forms, so far they are actually pretty trivial. Whats killing me, every time, is just trying to remember all the different ways to make complicated integrations work. I am getting better though, I can recognize that

S(1/1-x^2) is going to be some trig sub/identity

S(x/1-x^2) is going to be a u substitution

yada yada yada.

Still though, I totally underestimated how hard this would be.

 

[Hailey Knight]

Oh god why did I do this?

I could understand all this shit; I just don't wanna. :humph:

 

[tegid]

Oh god why did I do this?

@ Fade- yep. You can spend hours trying to get the analytical solution for some differential equation, or just put it into the appropriate form, jump into excel (or some higher dimensional program like matlab depending), set your dx or dt or whatever to small, and run it. Sadly though, that's not a correct answer for my diff eq prof. Numerical methods for the win.

Anyways, hit a problem I am having some trouble with, thought yall could help again. I'm getting a lot better at this, but I am still rusty on my integral calc. Right now I am messing with linear diff eqs. The general form is :

y' + p(x) y = q(x)


here's the problem

y' + y * cot x = csc^2 x
p(x) = cot x
h(x) = ln(sin x)
e^h(x) = sin x

Multiply the starting equation by e^h(x) and get

sin x[y' + y cot x] = sin x * csc^2 x
[y*sin x]' = csc x = sin ^-1 x

now this is where I am wondering if I have done it right, because the integral for that is not simple:

y*sin x = integral(csc x) = - ln |csc x + cot x| + C

y = sin x * -ln(csc x + cot x) + C sin x

There has to be a more simple form of that. Any thoughts?

Shit, I thought I had answered this, I'm sorry... If you look for int(1/sinx) instead of int(cscx) the tables will give you a simpler solution (ln(tan (x/2)) instead of ln(csc x + cot x))

 

[Necronic]

Oh god why did I do this?

ahhh good call. Thats been the hardest part of the class for me so far, not the dif eq forms like first order or seperable or whatever, but the basic calculus to manipulate those forms, worst of all being the trig identities. I'm really concerned about how many of the identities I will have to have memorized for the test. In the quizes so far I have seen stuff like the integral of an arctan or a tanh or something, which is a fucked thing to have to memorize.

 

[tegid]

Oh god why did I do this?

the hiperbolic functions can be separated into real exponentials, that's a way of doing the integrals without memorizing, I guess (I have always been allowed to take an immediate integral extended table to exams, except for one time many years ago, and I don't remember any of it)
(regular trig functions can be separated into exonentials too, but they are complex exponentials which makes it shitty)

 

[Necronic]

Oh god why did I do this?

I am really starting to get the hang of this, plus (and this is a big plus) I have the wonderful advantage of taking this class after the introduction of Wolfram-Alpha.com. For anyone that doesn't know of this site, its amazing for math. Its like a search engine for math/statistics. Front page is like google, just a bar to type stuff in. Type in 'integral of y/(y-x^2)' and it shows you the integration answer as well as all the steps to get it. Type in : y''-3y'+2y=0 it will identify it as a linear dif eq, then give you the general solution as well as the steps to get there. Type in your name, and you get all these cool stats on your name.

Honestly thought it may be too good. I think it will break the non-test side of the class to a degree.

 

[fade]

Oh god why did I do this?

Sometimes mathematica (which I'm guessing is the engine since it's wolfram) doesn't give the most simplified or direct answer. But it is helpful. If you still need help, I can look at your problems. I avoided them before because of time constraints (other than the silly joke).

 

[Necronic]

Oh god why did I do this?

Now I'm reading about Wolfram. You know, if this dude had been bitten by a radioactive spider or bombarded by gamma rays he would totally have a place in comics:

-Published his first paper on particle physics at 15
-published a very important paper on quarks at 18
-recieved a PhD from CalTech at 20.
-borderline polymath (expertise in: particle physics, cosmology, cellular automata, computational complexity theory, computer algebra, and developed the Mathematica program)

on the other hand, here's my CV
-slept for an average of 14 hours a day from the ages of 18-23
-graduated with a bachelors in Chemistry from university of texas at 25 (delayed 2 years due to library fees.)
-was able to get up before 8 AM for 2 work weeks at the age of 27
-at the age of 28 woke up on a half eaten hamburger. Considered eating it.
-watched the entire series 'the wire' in 3 weeks at the age of 28
-has had a couple of good ideas. Decided to sleep instead.
I have to say I am impressed with my accomplishments so far this year.

 

[Necronic]

Well, I just took my first test. And, how shall I put this. WTF!?!?!?!

The test was 10 questions long, 50 minutes. The first 6 were multiple choice, the next 4 were worked problems. The test was taken at a computing testing center.

Each of the multiple choice questions was something involving rearranging a general solution or a diff eq in order to get the other one. Or getting an orthogonal or whatever. This was the worst part. 5 minutes per question, and you have to finish with your equation in the same form as the correct choice. On thing I have learned with this kind of stuff is that there are many ways to skin a diff eq. In a couple of cases I worked the problem out, got my answer, and looked down and whamo. It wasn't there. So I just wasted 5 minutes on teh problem.

The worked problems weren't so hard, not nearly as bad. Those I was able to do, and I had the ease of knowing that if I missed a sign somewhere or if I didn't get the elegant solution I would still get some credit.

Honestly, who can work 10 complicated first order/bernouli/homogenous and/or second order equations in 50 minutes without missing a sign?

Good news was I ran into some students before the test that showed me how to work

y1=e^ax * cos(Bx)

backwards, and how to solve

y''+ay'x^-1+Byx^-2 = 0

with z substitution to get x^r

that saved my ass (although I'm sure I messed up a sign.....)

so, anyways, I am going to seriously have to ramp up my studies so I can do these integrations faster.