How Do You Solve a Second Order Linear ODE with Non-Constant Coefficients?

[pat666]

Homework Statement

[tex] 22e^{2t}=y''+8y'-95 [/tex]

Homework Equations

The Attempt at a Solution

I've been reading a textbook on this and think that I should use "method of undetermined coefficients" I know [tex] r(x)=ke^{\gamma*x}[/tex] so [tex]y_p(x)=Ce^{\gamma*x}[/tex]
The trouble is after reading the entire chapter I don't know what to do with that? I'm using Advanced Engineering Mathematics by Kreyszig and it is the worst textbook I have ever used there is huge gaps in methods and no easy to understand descriptions of the maths! Anyone know of any good textbooks for DE's??

Thanks for any help

 

[grey_earl]

I cannot recommand you a good book, but here is a guide to the solution of your particular problem.

First you should bring all y terms on one side and the rest on the other, so you have
[tex]y''(t) + 8 y'(t) = 22 \exp(2t)+95[/tex]. Now you see that there is no y(t) term, only derivatives, so you can substitute a new function z(t) = y'(t) to simplify. That's completely general up to now.

In your specific case you can guess the form of z(t) (it should have two unknown constants) and then substitute into the differential equation to determine those constants. A last integration then brings you to a specific (particular) solution y(t).

Since your differential equation is linear, you can add to this particular solution any solution of the homogeneous equation (the one where you get rid of the right hand side). Do you see why?

Try, and if you're still stuck, come back, show what you tried and ask for more :)

 

[SVXX]

Well, how I've been taught involves finding the complementary function and the particular integral and adding them to get the complete solution. I'd follow these steps -:

1. Write in terms of the D operator. Here it would be -: (D² + 8D)y = 22exp(t) + 95exp(0).
I wrote 95 as 95exp(0) so that I can have a recognizable particular integral for it.

2. Find the auxiliary equation and its roots. The auxiliary equation here would be -: D² + 8D = 0. Surely you can find the roots by quadratic concepts.

3. Now as the roots would be real and different here(as is obvious), if the roots be A and B, the complementary function is of the form -:
c₁e^Ax + c₂e^Bx + ... so on as many number of roots there are. Here there are two. The c's are arbitrary constants.

4. To find the particular integral, note these inverse D operator formulae as 1/f(D) of the RHS is the particular integral -:

This isn't multiplication, this is the application of the inverse D operator on the RHS to find the particular integral. Note that T is a function of t(T being the RHS, then). Our particular integral here would be of the form -:

5. This is getting long, but we're almost there. Now to get the PI of a function like e^(t), we have to put the value of the coefficient of t from e^(something t) into D and check if the expression evaluates to 0(which it may in this question). For example, if it were e^(t), I'd put 1 into D² + 8D. Operate on e^(2t) and e^(0t) separately and add their PIs.

6. IF it does evaluate to zero (and it might here ;) ), we differentiate the expression and then put it again to check. We keep doing this until it doesn't evaluate to zero. The cases where it evaluates to zero are called cases of failure. We may or may not require those inverse D formulae I wrote, depending on the question. Just thought you should know 'em, though. They do come in handy :)

In the end,
Complete Solution = Complementary Function + Particular Integral. It isn't as hard as it seems, its just hard to explain over the internet!
Btw, this is from my college-supplied book...I can't really pinpoint you to a good book as I don't know your syllabus.
Note : Simply differentiating the expression is done specifically in the case of e^(x) functions. If you want, PM me and I'll send you some of my notes for all 4 cases.

 

[hunt_mat]

Doesn't this have a direct first integral?

[tex]
11e^{2x}=y'+8y-95t
[/tex]
This is now solveable via the integrating factor method.

 

[pat666]

Hey thanks for the responses:
so far I have:
roots are 0 or -8
the complementary equation is
[tex] c_1e^{0x} + c_2e^{-8x}=C_1+c_2e^{-8x}[/tex]

Then [tex]1/(d^2+8D)(22e^{2t}+95) [/tex]
I am getting a bit confused here, would the coefficient you are talking about for e^(2t) be 1 or 2? none of them make D^2+8D=0 Also what does PI stand for/mean I know we've used it in these questions but I can't remember what exactly it means? also is a=A or where did a=-8D come from?

Thanks

 

[SVXX]

For e^(2t), the coefficient would be 2. e^(2t) won't make it zero, so just put 2 into D^2 + 8D and you get the particular integral as e^(2t) * 22/20. However, in 95e^(0t), it will be zero, so differentiate till it will be non zero on putting D = 0, OR until you get a constant after differentiation. PI means particular integral.
Those inverse D formulae are not required for this case!

 

[HallsofIvy]

Note that your equation can be written
[tex]y''+ 8y'= 2e^{2t}+ 95[/tex]
so its characteristic equation is [itex]r^2+ 8r= 0[/itex] which is easy to solve. Since r= 0 is a root of that equation, so a constant is already a solution to the associated homogeneous equation, you then seek a particular solution of the form [itex]y(t)= Ae^{2t}+ Bt[/itex]

[itex]y'= 2Ae^{2t}+ B[/itex] and [itex]y''= 4Ae^{2t}[/itex]. Put those into the differential equation and solve for A and B.

 

[pat666]

r^2+ 8r[= 0/itex] which is easy to solve. Since r= 0 is a root of that equation, so a constant is already a solution to the associated homogeneous equation, you then seek a particular solution of the form [itex]y(t)= Ae^{2t}+ Bt

only posted because the original didn't show up in my browser

 

[epenguin]

See also https://www.physicsforums.com/showpost.php?p=3206657&postcount=7

There I am talking about particular integrals and assuming known:

Well, how I've been taught involves finding the complementary function and the particular integral and adding them to get the complete solution.

In the end,
Complete Solution = Complementary Function + Particular Integral. It isn't as hard as it seems, its just hard to explain over the internet!

 

[HallsofIvy]

r^2+ 8r[= 0/itex] which is easy to solve. Since r= 0 is a root of that equation, so a constant is already a solution to the associated homogeneous equation, you then seek a particular solution of the form [itex]y(t)= Ae^{2t}+ Bt

only posted because the original didn't show up in my browser

Thanks. I've fixed my post. Forgot a "["!

 

[pat666]

Hey guys,
I have managed to really confuse myself now!(
HallsofIvy you said to sub y' and y'' to solve for A and B which yields
[tex]4A*e^{2t} +8(2Ae^{2t}+B)=2e^{2t}+ 95[/tex] But my problem here is that i have 3 unknowns but 1 equation? Is A=0 and B=-8 from svxx not valid for some reason? I also have confused myself so much that I can't see a linear sequence of events that have led me anywhere, I seem to be back at the beginning?
Not complaining, very grateful for the help from everyone!

thanks