# Multi-Step linear systems problem - Build ODE, impulse response, systems balance

1. Oct 21, 2009

### dankaroll

I messed up on the title of the thread.. I had a few questions but ended up solving them on my own. The only question I have is how to solve this ODE. Thanks.

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

Solve the ODE

(dx/dt)= f(t) - .1x

Where f(t) = 2+sin(t)

2. The attempt at a solution

I faintly remember doing these types of ODE's in diff eq class, but there was always a x/t term so it ended up being a nice equation. Not really sure how to approach one of these though.

2. Oct 21, 2009

### rey242

I learned this recently in my class. Try moving all the x factors to the left and see if you can do something there.

3. Oct 21, 2009

### dankaroll

I think its coming back to me.. I need to find an integrating factor

I found that the integrating factor would be e-.1t

If I multiply both sides by it and integrate I end up with:

e-.1tx = -20e-.1t+int(sin(t)e-.1tdt)

So I end up having the int(sin(t)e-.1tdt)

Which isn't being too cool with letting me solve it.. any suggestions?

4. Oct 21, 2009

### rey242

You're on the right track.
Remember you have a dx/dt on the left so the equation would be
(dx/dt)+.1x= 2+sin(t)

by turning this into a different notation you get:
x'+.1x=2+sin(t)

So then you can now multiply by the integrating factor and get
e^(.1t)x'+.1e^(.1t)x=2+sin(t)

Does the left side look familiar?

5. Oct 21, 2009

### Dick

Yes, I think it's coming back to you. I would say the integrating factor is for x'+x/10 is exp(t/10) though. Keep working on it. I think it will come back. To integrate sin(t)*exp(t/10) you integrate by parts. Twice. Is the memory coming back?

6. Oct 21, 2009

### dankaroll

As of right now I'm still stumped.. but I did realize I made a mistake with a sign.

After integrating twice I end with:

et/10x = 20et/10 + 10et/10sin(t) - 100et/10cos(t) - 100 $$\int$$et/10sin(t)dt

I think I might see something..

$$\int$$et/10sin(t)dt = et/10x-20et/10?

Still a noob with the symbols :P

Thanks for all your help so far!

7. Oct 21, 2009

### rey242

Try my way...
e^(.1t)x'+.1e^(.1t)x=2+sin(t)

The left side is just the product rule. So if you do this:
d/dt(e^(.1t)*x)=2+sin(t)

Then you can do a really simple integration. But you could also do it Dick's way.
:)

8. Oct 21, 2009

### dankaroll

I'm all for the easy way (dont shoot me!) but what I don't see in your way is why you only multiplied one side by the integrating factor. Should you multiply the other side as well?

9. Oct 21, 2009

### Dick

No, what's an x doing in there? You have to integrate exp(t/10)*sin(t)dt by parts using d(-cos(t))=(sin(t))dt. Then you integrate by parts AGAIN and you get the integral of exp(t/10)*sin(t) back again. But with a different sign. Solve for the integral of exp(t/10)*sin(t). I think you've probably seen this before.

10. Oct 21, 2009

### rey242

Yeah sorry about that, I forgot.
Multiply both sides by exp(.1t).

The integration is harder now but its still the basic concept.

11. Oct 21, 2009

### dankaroll

So I redid my integration and got..

et/10x = 20et/10 - et/10cos(t) - (1/10)et/10sin(t) - (1/100) $$\int$$et/10sin(t)dt

cmon brain dont fail me now.. i still dont see it arg

12. Oct 21, 2009

### rey242

Remember Integration by Parts has "tricks"?

$$\int$$ exp(t/10)sin(t)dt =20et/10 - et/10cos(t) - (1/10)et/10sin(t) - (1/100)$$\int$$ et/10sin(t)dt

13. Oct 22, 2009

### dankaroll

(I made a mistake with the 1/10 term, should be +)

anyways

OK!! I think I figured it out..

You can sub the integral parts for a variable.. in this case i'll call it Y

So I solve for Y, plug it back in where the integral was.

for y I ended up with

Y = (100/101)(20et/10 - et/10cos(t) + (1/10)et/10sin(t))

So then u plug it back into where the integral was, exp functions end up canceling out and u get your answer for x

right?

ah, what a complicated answer for such a simple looking problem. math is the master of disguise :)

Last edited: Oct 22, 2009