/sci/ - Science & Math

[OP !!yzPI6QwWJCg No.4821129]

Hello /sci/,

I'm currently struggling with understanding a solving method for linear 2nd order differential equations, and I hope someone could help.

The problem I'm trying to solve is

<span class="math">\displaystyle y'' - 3y' + 2y = e^{17x}[/spoiler]

The general idea is, if I'm not mistaken, to solve the characteristic equation first, since the solution of the inhomogeneous equation is equal to the sum of the general solution of the homogeneous equation and one particular solution of the inhomogeneous one.

In this case, the characteristic equation would be

<span class="math">\displaystyle \lambda^2 - 3\lambda + 2 = 0 [/spoiler]

which leads to the solutions

<span class="math">\displaystyle \lambda_1 = 1, \lambda_2 = 2 [/spoiler]

and the general solution of the homogeneous equation is

<span class="math"> \displaystyle y = A e^x + B e^{2x} [/spoiler]

Then, my notes say that, if <span class="math">\displaystyle y_p [/spoiler] is a particular solution, then it could be described as

<span class="math"> \displaystyle y_p = e^{cx} * x^r (P_n(x) cos(bx) + Q_n(x) sin(bx)) [/spoiler]

where <span class="math"> \displaystyle P_n(x), Q_n(x) [/spoiler] are polynomials, and <span class="math"> \displaystyle b, c , r [/spoiler] are real numbers.

Also it was said that <span class="math"> \displaystyle c + ib [/spoiler] is an r-fold solution of the characteristic equation.

That last sentence seems to be my main problem of understanding - what exactly do I do with it and why is it that way?

[OP !!yzPI6QwWJCg No.4821194]

Bump.

[Anonymous No.4821229]

I don't know what you're trying to do in the last part of your post, but whatever it is it's definitely over-complicated.

Just use <span class="math">y_p=\frac{1}{270}\mathrm{e}^{17x}[/spoiler] as a particular solution.

[Anonymous No.4821241]

Not sure what you're trying to do...

Just use <span class="math">y_p=Ae^{17x}[/spoiler] as a particular solution for some constant A, plug it into your diff. eq. and solve for A.

You'll get 17^2A-3*17A+2A=1.

[OP !!yzPI6QwWJCg No.4821262]

>>4821241
I hope I'd know what exactly my professor meant by it, my notes on that topic are a bit cryptic.

I'll try your idea, though.

[Anonymous No.4821269]

I'd say finding particular solutions are a matter of mathematical wit. Yes, the general formula is there, but you have to think before resorting to the very principles.

[Anonymous No.4821352]

This kind of stuff is better done with Laplace transforms. They make the whole thing pretty trivial.

[Anonymous No.4821374]

>>4821352

It's more suitable for initial value problems, if you want to find the general solution and apply transform you will have to carry y'(0) and y(0) around because you dont know their value, but in this case looks easy

[Anonymous No.4821385]

>>4821374
No matter what you will always have two parameters to carry around, call them A and B as in OP, or call them y(0) and y'(0), there is no difference. The family of solutions of a second order ODE will always depend upon them one way or the other.

[Anonymous No.4821396]

Damn it, OP.

I was so bothered that I didn't remember how to solve this problem that I went and grabbed my copy of Essential Mathematical Methods for the Physical Sciences by Riley & Hobson. Unfortunately, I couldn't find what I was looking for, since it's been too long since I used the book, and I can't remember all the proper terminology. I tried looking for your method with the P(x) and Q(x) polynomials, and couldn't find it, although it looks so familiar that it must be in there somewhere...

Honestly, I should have just gone straight to Wolfram Alpha, since it's more than adequate for problems like this one.
http://www.wolframalpha.com/input/?i=y%27%27-3y%27%2B2y%3Dexp%2817x%29
Click the "show steps" button next to the solution.

Anyway, your basic approach is correct, since you have an inhomogeneous linear differential equation with constant coefficients. Your complimentary function is definitely right.

To find the particular solution, I think you just have to pick a function that's a multiple of the right-hand side of your original equation:

y_p = C*exp(17x)

Substitute this into the original equation and you get

17*17*C*exp(17x) - 3*17*C*exp(17x) + 2*C*e^(17x) = exp(17x)
240*C*exp(17x) = exp(17x)
C = 1/240

So the particular solution is
y_p = (1/240)*exp(17x)

The general solution will be your complimentary function y_c plus the particular solution y_p.

y = A*exp(x) + B*exp(2x) + (1/240)*exp(17x)

That stuff you wrote with the P(x) and Q(x) is definitely something I've seen before; it might be a more general way of doing the problem, or it might not be applicable to this exact problem at all. I just can't remember; it hasn't been very long since I did differential equations, but I was never very good at it anyway.

[Anonymous No.4821421]

>>4821396

... Okay, wait, I just remembered what those polynomials are for.

When finding the particular solution, if the right-hand side of your differential equation is of the form a*exp(r*x), you'll want to pick the trial function b*exp(r*x). This is the case we have in your problem, which is why my trial function was y_p = C*exp(17x).

But if the right-hand side is of the form a*sin(r*x) + b*cos(r*x), you pick the trial function c*sin(r*x) + d*cos(r*x).

And if the right-hand side is a polynomial, your trial function should be a polynomial of the same order.

Finally, if the right-hand side is a product of any of the above, the trial function should be a product of the above trial functions.

So if the right-hand side of your equation were something like (x^4)*exp(2x)*sin(3x) -- in other words, a fourth-order polynomial times an exponential times a trigonometric function -- then your trial function would have to be a product of all these forms. In this case, you would use

y_p = exp(c*x) * x^r * [P(x)*cos(b*x) + Q(x)*sin(b*x)

as you wrote, and P(x) and Q(x) would be fourth-order polynomials.

So it's a more general way of finding the solution, but you don't have to use it all the time.

[OP !!yzPI6QwWJCg No.4821464]

>>4821421
Thank you for putting all that effort in it, even though the motivation wasn't all that selfless (hah!).
What you say makes a lot of sense to me, since you can just eliminate the terms you don't need in case the right hand side of the equation doesn't consist of them.

[Anonymous No.4821517]

>>4821464

No problem. I should mention, though, that in

y_p = exp(c*x) * x^r * [P(x)*cos(b*x) + Q(x)*sin(b*x)

I think the x^r is only there in case your particular solution would otherwise contain terms that already appear in the complementary function.

Let's say the right-hand side of your equation is a product of a polynomial, an exponential, and a trigonometric function. The polynomial component is represented by P(x) and Q(x), the trigonometric component is represented by sin and cos, and the exponential is there too. However -- and I'm going to quote the aforementioned textbook here -- "this method fails if any term in the assumed trial function is also contained within the complementary function y_c(x). In such a case the trial function should be multiplied by the smallest integer power of x such that it will then contain no term that already appears in teh complementary function."

So I think that explains the factor of x^r that was thrown into the particular solution in your original post.