Encountering "unsolvable equation" during simple ODE problem

[Larry Cosner]

Homework Statement

{Prob #27, Section 2.1 "EDE" (Boyce/Prima, 10thEd), pp. 40}

"Consider the initial value problem [and] find the coordinates for the first local maximum point of the solution, t>0."

Homework Equations

y' + (1/2)y = 2cost y(0) = -1

The Attempt at a Solution

I began this thinking it would be fairly straightforward, and readily found the general solution:

y = (4/5)(cost + 2sint) + ce^(-t/2)

To find "c" (ie: the specific solution) I used the initial values and got:

y=(4/5)(cost + 2sint) - (9/5)e^(-t/2)

In attempting to find the (first) local maxima -- making the assumption that it would be at a point where y' = 0 -- I took the first equation, which reduces to:

(1/2)y = 2cost ----> y = 4cost

and tried solving for t [by inserting my above-noted specific equation for y(t)]. Which ultimately brings me to the equation:

9e^(-t/2) = 8sint - 16cost

Now I should note that [since the answer, meaning the point value (1.364312, 0.820082) is in the back of the book] all my equations balance. So I assume I've done the various derivations and integrals correctly. My general and specific solutions seem correct. But I cannot for the LIFE of me think of how to simplify the equation above to some form where "t" can be solve for. (Though as I mentioned, putting the 'known' value in does balance.)

Logically I'm overlooking something that should either allow me to simplify that equation, or I've created the wrong equation. But I'm stuck.

[FWIW - I'm not formally taking a class; I'm considering a project that requires re-building of my DiffyQ knowledge, now 30ish years stale. So I recently worked through the most current version of Thomas' Calculus and am now working through this (introductory) ODE text. If I had an instructor or TA grad student, I'd be bugging one of them. But I live in a town sans 4-year college, so I can't think who to 'bug'! Thanks for considering help!]

 

[RUber]

I normally like to solve this type of problem numerically.
However, by using the half/double angle formula you might be able to simplify it to at least one variable--substitute $\theta = t/2:$
$9 e^{-\theta} = 16 \sin \theta \cos \theta - 16 (\cos^2 \theta - \sin^2 \theta)$
I do not clearly see a quick solution from there...but I feel like it is a step in the right direction.

 

[SteamKing]

In attempting to find the (first) local maxima -- making the assumption that it would be at a point where y' = 0 -- I took the first equation, which reduces to:

(1/2)y = 2cost ----> y = 4cost

and tried solving for t [by inserting my above-noted specific equation for y(t)]. Which ultimately brings me to the equation:

9e^(-t/2) = 8sint - 16cost

Now I should note that [since the answer, meaning the point value (1.364312, 0.820082) is in the back of the book] all my equations balance. So I assume I've done the various derivations and integrals correctly. My general and specific solutions seem correct. But I cannot for the LIFE of me think of how to simplify the equation above to some form where "t" can be solve for. (Though as I mentioned, putting the 'known' value in does balance.)

Logically I'm overlooking something that should either allow me to simplify that equation, or I've created the wrong equation. But I'm stuck.

You're overlooking the fact that this equation you're trying to solve here is not an algebraic equation. You can't manipulate it like a quadratic polynomial, for example, and solve for the unknown variable.

Equations involving logs and exponentials are examples of transcendental equations. Typically, they are solved either by using a numerical technique, like Newton's method, or graphically, by plotting the function.

 

[Ray Vickson]

Homework Statement

{Prob #27, Section 2.1 "EDE" (Boyce/Prima, 10thEd), pp. 40}

"Consider the initial value problem [and] find the coordinates for the first local maximum point of the solution, t>0."

Homework Equations

y' + (1/2)y = 2cost y(0) = -1

The Attempt at a Solution

I began this thinking it would be fairly straightforward, and readily found the general solution:

y = (4/5)(cost + 2sint) + ce^(-t/2)

To find "c" (ie: the specific solution) I used the initial values and got:

y=(4/5)(cost + 2sint) - (9/5)e^(-t/2)

In attempting to find the (first) local maxima -- making the assumption that it would be at a point where y' = 0 -- I took the first equation, which reduces to:

(1/2)y = 2cost ----> y = 4cost

and tried solving for t [by inserting my above-noted specific equation for y(t)]. Which ultimately brings me to the equation:

9e^(-t/2) = 8sint - 16cost

Now I should note that [since the answer, meaning the point value (1.364312, 0.820082) is in the back of the book] all my equations balance. So I assume I've done the various derivations and integrals correctly. My general and specific solutions seem correct. But I cannot for the LIFE of me think of how to simplify the equation above to some form where "t" can be solve for. (Though as I mentioned, putting the 'known' value in does balance.)

Logically I'm overlooking something that should either allow me to simplify that equation, or I've created the wrong equation. But I'm stuck.

[FWIW - I'm not formally taking a class; I'm considering a project that requires re-building of my DiffyQ knowledge, now 30ish years stale. So I recently worked through the most current version of Thomas' Calculus and am now working through this (introductory) ODE text. If I had an instructor or TA grad student, I'd be bugging one of them. But I live in a town sans 4-year college, so I can't think who to 'bug'! Thanks for considering help!]

Ultimately, you must resort to a numerical solution method. You can write your equation as $f(t) = 0$, where $f(t) =16 \cos(t)-8\sin(t) + 9 e^{-t/2}$, then try a host of numerical equation-solving methods, such as Newton's method, bisecting search, the secant method, etc.---I am giving you names you can use as Google keykwords. Basically, though, you should always start off by plotting f(t) over some range [0, T] (if you can) to get a rough idea where the root lies. Then you can home in on more accurate estimates using one of the methods above. Alternatively, you can submit the problem to Wolfram Alpha (a free version of Mathematica lite) or one of the other available on-line equation solvers. If you have the spreadsheet EXCEL (or equivalent) you can use the Solver tool to get a solution.

In this particular case you can simplify the problem somewhat, but that does not really allow for a solution---only for a possibly simpler and possibly more accurate analysis. Use the trigonometric identity $\cos(t+w) = \cos(w) \cos(t) - \sin(w) \sin(t)$ to write
$$16 \cos(t) - 8 \sin(t) + 9 e^{-t/2} = r \cos(t+w) + 9 e^{-t/2}, \\ \rm{where}\;\; r = \sqrt{16^2+8^2}, w = \arccos(16/r)$$
Putting $t+w = y$ we end up with the equation
$$0 = r \cos(y) + 9 e^{w/2} e^{-y/2} = 17.88854382\, \cos(y) + 11.34807790\: e^{-y/2}\,.$$

 

[Larry Cosner]

Thank you to all who replied; I had begun to suspect -- but lacked the mathematic knowledge/terminology to articulate -- that the equation was, by definition, and unsolvable. So explaining the fact, and why, was VERY helpful! (I sort of wish that the text-book authors would have indicated -- in this problem specifically -- the expected need for a numeric approximation solution, But then, I suppose you don't get this kind of warning in Real Life. So I'd best get used to it ... <sigh>.

Thanks again! I can now move forward with my life and not sit huddled under my desk, whining, with my nurse saying: "No, really, you have to come out and see patients, stupid!".

 

[rude man]

"Real life"? Easy! Use Wolfram Alpha!
Made your day, or what?!

 

[Larry Cosner]

"Real life"? Easy! Use Wolfram Alpha! :)
Made your day, or what?!

Funny you should mention this. Found out a couple of days ago that I was eligible (through an association between my hospital and the local Community College) to purchase Version 10. So, a few minutes, a bit of typing, some electrons running about the Internet (sort of), and me and my credit card are now proud owners! I'm already having fun with it. So thanks for adding fuel to the fire of my joy. (That may not be my best metaphor.)