# Homework Help: 2nd Order Differential Equation Problem

1. Nov 19, 2009

### jonnejon

1. The problem statement, all variables and given/known data

Let y be the solution of the initial value problem:
y'' + 2y' + 2y = 0 , y(0) = 0 , y'(0) = 5

The maximum value of y over 0 less than or equal to x less than infinity is ??.

3. The attempt at a solution
r = -2 +/- i

I solved it:

y = c1 e^-2x cos(x) + c2 e^-2x sin(x)
y' = -2c1 e^-2x cos(x) - c1 e^-2x sin(x) - 2 c2 e^-2x sin(x) + c2 e^-2x cos(x)

y = c1 = 0
y' = -2c1 + c2 = 5 => c2 = 5

y = 5 e^-2t sin(x)

But I am terrible at trig. What is the maximum value of y? I got (x = -3pi/2 so therefore max y is 5 e^(6pi/2) but that is incorrect. I have one more attempt and don't want to lose out on this question.

Thanks.

Last edited: Nov 19, 2009
2. Nov 19, 2009

### Staff: Mentor

The roots of your characteristic equation are wrong. They should be r = -1 +/- i.

3. Nov 19, 2009

### jonnejon

Yeah, I missed that but doesn't change my answer much.

r = -1 +/- i

I solved it:

y = c1 e^-x cos(x) + c2 e^-x sin(x)
y' = -c1 e^-x cos(x) - c1 e^-x sin(x) -c2 e^-x sin(x) + c2 e^-x cos(x)

y = c1 = 0
y' = -c1 + c2 = 5 => c2 = 5

y = 5 e^-x sin(x)

But I still can't find the max value of y. Help please?

I think max of y is 5 e^(3pi/2) where x = -3pi/2. I don't want to exhaust my last attempt. Any confirmations?

4. Nov 19, 2009

### Staff: Mentor

How do you normally find the maximum value of a function? You probably spent the good part of a quarter or semester doing that.

5. Nov 19, 2009

### jonnejon

I really don't know, my professor moves way to fast and doesn't give enough examples. I also completely forgot about that. If you show me how to do it instead of giving hints, I am sure I will learn it better that way. I would really appreciate that. Thanks.

6. Nov 19, 2009

### Staff: Mentor

And I'm sure that you will learn it better if you dig for it than if I just tell you. Crack your calculus book open and see if it has anything about finding the largest value of a function.

7. Nov 19, 2009

### jonnejon

I am sorry, but I really don't know. I have been looking for awhile and I can't seem to find it. Did I do the differential equation correct though?

8. Nov 19, 2009

### Staff: Mentor

Yes, the solution of the initial value problem is y(x) = 5e-xsin(x).

Any maximum or minimum will come at an endpoint of the interval in question or at a point at which the derivative is zero. I truly hope that rings a bell with you.

9. Nov 20, 2009

### jonnejon

Yeah, I found out that you set the derivative to zero but as I said before I am terrible at trigonmetry. So:

y'=5e^-x (cos(x)- sin(x))
0=5e^-x (cos(x)- sin(x))
5e^-x cos(x) = 5e^-x sin(x)
cos(x) = sin(x)

Whats next?

10. Nov 20, 2009

### Chewy0087

try solving for x? (dividing throughout by cos(x))

11. Nov 20, 2009

### jonnejon

So, 0 = sin(x)/cos(x) = tan(x)? so tan (0)=0 so the max value of y is 0???

12. Nov 20, 2009

### Chewy0087

cos(x)/cos(x) does not equal 0....

what is 1/1, 5/5, pi/pi ?

13. Nov 20, 2009

### jonnejon

Oh yeah, did the algebra wrong...

So, sinx/cosx = 1 = tanx
x=45?

so I should input the answer as:
y = 5 e^(-45) sin(45)?

14. Nov 20, 2009

### Chewy0087

i think you need to put it in radians, but your method is correct

15. Nov 20, 2009

### jonnejon

Yes!!

I got it. Thanks Mark and Chewy. Yeah, I forgot to set my calculator back to radians, thanks for the help guys.

16. Nov 20, 2009

### Staff: Mentor

There are an infinite number of solutions to the equation tan x = 1 on the interval [0, infinity). You have found only one of them. All you know is that dy/dx = 0 at x = pi/4. You don't know whether you have a maximum, minimum, or neither at this point.

17. Nov 20, 2009

### Staff: Mentor

You've mentioned a couple of times that you're terrible at trigonometry, you seem to have some trouble with algebra, and you didn't remember that the derivative of a function could be used to find the maximum and minimum values of that function. Now you are apparently in a class in differential equations, where knowledge of these areas is assumed. To be weak in all these areas is definitely not a recipe for success in differential equations.

I don't know what classes you are planning to take in the future, but you probably are going to have a much harder time of it until you can get up to speed in the areas where you are weak. I can't imagine that you are having an easy time of it in your present class.