Yeah for my applied mathematics class, I have to find explicit solutions which include the domain and range , plus all assymptomoes within the range.
But yeah I used newtons method and it gave me that answer thanks.
I tried to solve a fourth degree polynomial by hand for a Field theory course, spent a ton of time and I probably messed up a complex square root and had to repeat everything.
never again.
God damn that sounds terrible and I ferreting at the same time.
I think I had to do something similar to that while taking the FE/EIT Test and while I didn’t remember and didn’t care to remember, I just guessed. If that number was too low, then I would just take the opposite, the negative or positive of it, then try that, if not just average the two.
First time I had to do something like this was in an abstract algebra class. As you’ve seen, formulas for solutions to 3rd and 4th degree polynomials are quite involved. There isn’t an explicit formula that can solve a 5th degree polynomial in general, so I wouldn’t be too bummed about it. Sometimes you can get lucky and the Rational Root test will net you a root of a polynomial, but that’s why numerical methods are so useful.
so, I’m doing second order differential equations, and we’re doing unknown coefficient technique.
y"+2y’+y=2e^(-t) and complementary general solution is
y = c_1e^(-t) + c_2t*e^(-t) + y_p (can anybody also tell me why a term is multiplied by t?)
this is one of those that can’t be solved under normal circumstances using this method.
when our model is, Ae^(-t) = y_p, and then y_p’ and y_p" is plugged back into original eqution, it doesn’t equal zero and the model shares a term with the general solution. At, Ate^(-t), it cancels into 0=2, and even then it cannot be used becuase it again shares the term on the right hand side.
at A*(t^2)*e^(-t),
y_p" + 2(y_p’) + y = 2e^(-t), it sums down to
4A -A*t^2 = 2,
this is more of an algebra question, but , how am I supposed to find what the unknown coefficient is, when I have two unknowns and only one equation…
Am I doing something wrong? If I solve for A, and I plug it back in, 2=2. I tried solving it by assuming that, at what values of T would the equation = 2,
the coefficient A cancel, and I’m left with, t = 2+2^.5 and t = 2-2^.5. Now I think it’s a huge jump for me to assume, that, at any of these two values, I can find the value of my unknown coefficient…which means A = 1.
Now, it worries me more because the solution is
y = c_1e^(-t) + c_2t*e^(-t) + (t^2)*e^(-t)
I don’t want to make assumptions like that.
Don’t you get the unknown coefficients from boundary conditions usually? Or initial conditions in this case, since you’re dealing with functions of time.
You’re doing second order differential equations, so you’ll need two initial conditions to solve the problem. Also, your homogeneous solution will be two exponential terms (like what your model is, but I’m not sure why the second exponential term is multiplied by t. Probably a typo?), each with their own base and exponent coefficients, which is why you will be needing two initial conditions to algebraically solve for two unknown base coefficients with a system of two equations.
IIRC, you figure out what C1 and C2 are by when the diif eq is zero and such. =P
No it’s not a typo. The book has the solution listed at that, and it’s basic example of this type (where your specific solution model is multiplied by t until no terms are shared between specific solution and the general solution, And when you plug in the models derivatives into the original equation back in, does sum to something like 0=2)
I also didn’t get any values, just a differential equation asking for a general solution, and the specific solution to the non homogenous part.
that’s for homogenous equations. This is non homogenous equations, where there are two solution. A linear combination of the homogenous equations and non homogenous part.
Y " c1y1 + c2y2 + yp
Where yp is built of using the g (t) of
Y"+y’+y=g(t)
Yeah, the solution for non homogenous equations is homogenous solution + particular solution, right? So you get c1 and c2 from the homogenous equation, because they are part of the homogenous solution. Meaning yes, you get c1 and c2 from when the left hand side is equal to zero, because that’s the homogenous equation.
Hmm. If you don’t have values, just use yo (lol idk how to type y-nought) and y’o. But looking at your post again, your problem isn’t really how to get c1 and c2, right? Since you’re trying to get the particular solution?
From the looks of it what you did was right already. The final model for yp you arrived at was At^2exponential. Plugging it into the DE was right, since if your y is just homogenous solution + particular solution, the derivatives will also just be (homogenous solution)’ + (particular solution)’ but we know the DE for y=homogenous solution is just 0 by definition.
Could you re-check if you did the summing down of the DE correctly? Does it really sum down to 4A-A*t^2=2?
Because I think the step after substituting yp into DE is you compare coefficients: constant of left hand side should equal constant of right hand side, coefficient of t of LHS should equal coefficient of t of RHS, and so on. But looking at 4A-A*t^2=2, A should equal 2 (and 0 lol), but we know that’s not right since in your book A=1 (and 2 is not equal to 0 lol).
EDIT: ah fuck, I understand why the c2 term has t multiplied to it now. Remember that a second order DE of the form ay’’+by’+cy=f(t) is a damped oscillator with a forcing function? The initial assumption you make about the solution is that it’s of the form Ae^(alphat) right? Then you solve for alpha from the homogenous equation and you’ll end up with a quadratic equation in alpha.
There are three cases now: (1) there is exactly one real root for alpha (critically damped, discriminant is positive and a perfect square), (2) there are two real roots for alpha (overdamped, discriminant is positive), (3) there are two imaginary roots for alpha (underdamped, discriminant is negative).
Depending on the alpha you get, you will re-model your solution for y.
Case 1 (critically damped): you assume the solution is (A+B*t)e^(alphat), since you only have one alpha but need two exponential terms.
Cases 2 and 3 (overdamped and underdamped): you assume the solution is A_1e^(alpha_1t) + A_2e^(alpha_2t). You have two different As (A_1 and A_2) because you have two roots for alpha. The only difference between cases 2 and 3 is that case 2 is a real exponential, so you just take y to be an exponential function, while case 3 is an imaginary exponential so you take y to be a linear combination of sines and cosines.
It seems your book assumes critical damping which is why it uses (c1+c2*t)*e^(-t). Which is kinda weird lol 'cause most generalizations of damped oscillators assume underdamping because the solutions are sines and cosines which is what we are familiar with when dealing with oscillators.
I checked it and that’s what it adds up to…Even the ti-89 agrees
Unless I’m supposed to do y _p as
(At^2+Bt+C)e(^-t)
But in my book, and online resources, I didn’t see that they did This, and I doubt this is going to work. I’ll try this out tomorrow anyway as an exercise in futility
Thanks for the explanation. It seemed familiar, and I know remember this being taught in introduction to circuits.
Wait, are you sure? I did it myself and I’m getting 2A = 2 in the end.
Could you check if I’m doing the right thing:
DE is y’’ + 2y’ + y = 2*e^(-t)
The y we will use is y_p, which we are modeling to be At^2e^(-t)
y_p’ = 2Ate^(-t) - At^2*e^(-t)
2y_p’ = 4Ate^(-t) - 2At^2*e^(-t)
y_p’’= 2Ae^(-t) - 2Ate^(-t) - 2Ate^(-t) + At^2e^(-t)
Substitute all these into the DE, and divide by e^(-t) so it’s easier to type:
[2A - 2At - 2At + At^2] + [4At - 2At^2] + [At^2] = 2
[2A - 4At + At^2] + [4At - 2At^2] + [At^2] = 2
Rearranging to go from highest order in t term to lowest order in t:
At^2 - 2At^2 + At^2 - 4At + 4At + 2A = 2
2A = 2
A = 1
Your y_p, then, is 1t^2e^(-t) or simply t^2*e^(-t).
So your final complete solution for the DE is
y = (c_1 + c_2*t)*e^(-t) + (t^2)*e^(-t)
which is what your book says is the answer.
Could you go over everything? I don’t have paper and pen at hand so I did all this in notepad. I might have overlooked something because of the notation.
that’s correct, for some reason, I had my double derivateve come out as zero when I did it by hand, and throught the ti-89. I need to slow down. It’s always costing me points on all my exams.
thank you much.
Yeah sure. Btw this was a lucky case where the t and t^2 terms cancelled out and you were left with 2A=2.
Your original question was how to get A if you have two unknowns and just one equation, right?
You have to compare coefficients in that case. In the problem, t is an independent variable so it can take on any value. If your LHS has a t term but the RHS is a constant, you know that the t term has to have a zero coefficient so that the varying of t will have no effect on the value of the LHS expression, because the LHS has to be constant.
So basically just look at the constant terms of the LHS and RHS and equate them, so that’s one equation. Look at the coefficients of the t terms of the LHS and RHS and equate them, so now you have two equations. Look at the coefficients of t^2 terms of the LHS and RHS and equate them, so now you have three equations. That’s how you solve for multiple unknowns in this case.
Note that you can only compare coefficients of unknown variables, like t in this case. In this case, A is unknown but it’s a constant by the nature of the problem and your modeled solution, so you can’t compare coefficients of A.
Can somebody give the sequence in which I should learn mathematics?
Elememtary Arithmitic -» algebra -» geometry -» intermediate algebra -» trig -» calculus
I want to teach myself more , especially since linear algebra has been helpful. Even though I failed it. Hahah
You should know all those subject post high school.
But if you ever study Analysis (I think that’s what it’s called) you’ll probably go back and ask why or what was that again?
I do know them. It was an example of what I meant by sequence.
I’ve taken thus far for college
trig -> calculus -> linear algebra -> differentials
I’m trying to fit the second sequence of differentials which is Fourier series, partial differential equations, etc, into my schedule.
However, I want to teach myself higher level math, so I’d like to know what topics should I learn first, and what topics come after. Linear Algebra has really helped me understand at a basic level why some of the things I’m learning in differentials work, helps me a lot. So I’d like to know how to tackle math as an independent subject.
If you’re serious, then you should go back to your high school sequence you described and ensure you have mastered that first.
Without algebra, you cannot succeed, so you have to be able to do that without even thinking twice about it. You have to be comfortable with Inequalities too.
You also have to have mastered geometry. You should be able to find any length/angle/segment, etc. that you want or would want.
Furthermore, geometry should have been your first intro to proof. So, you should be comfortable reading and writing geometric proofs as needed. Look at the proof in Stewart’s Calculus text where he derives the limit of sin(theta)/theta = 1 as theta goes to zero. You should be able to understand what he did and why he did each step.
Geometry is important because trig is right angle geometry. You should know the unit circle like you know the alphabet. You should also be comfortable swapping trig identities as needed.
Algebra, geometry and trig are all important for calculus because calc is analytic geometry and certain sets of integrals require trig substitution. You also need to be comfortable with inequalities because you can’t use the definition of a limit or the squeeze theorem, among other things, with out them.
So before you study higher math, you should master high school math because your skill in the latter dictates your skill in the former.
You should also study propositional logic because higher math is more or less all proofs.
Do those things if you’re serious
I should also add that this list is in no way exhaustive. There are dozens of other things you would need to do.
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