# Intergrating Factors

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Mar 15th 2010, 11:22 AM
jamm89
Intergrating Factors
Been struggling to get started on this for hours.
The temperature of an object at time t is found from the equation
$\frac{d\theta}{dt}$+ 0.1θ = 5 - 2.5t
Given that when t=0, temperature is 60˚C. find the temperature when t=2 seconds
Any advice on getting started much appreciated.
Thanks
• Mar 15th 2010, 12:02 PM
tom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/asy/intFactor/simple.png

... where the chain rule...

http://www.ballooncalculus.org/asy/chain.png

... is wrapped inside the product rule...

http://www.ballooncalculus.org/asy/prod.png

... (legs un-crossed version). Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Now fill out the rest of the bottom row, and you'll have an exact product-rule derivative on the left-hand side, and you can integrate both sides with respect to t.

The general drift is...
http://www.ballooncalculus.org/asy/maps/factor1.png

And the rest...

_________________________________________
Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
• Mar 15th 2010, 12:03 PM
Miss
Quote:

Originally Posted by jamm89
Been struggling to get started on this for hours.
The temperature of an object at time t is found from the equation
$\frac{d\theta}{dt}$+ 0.1θ = 5 - 2.5t
Given that when t=0, temperature is 60˚C. find the temperature when t=2 seconds
Any advice on getting started much appreciated.
Thanks

You know you can solve this by the integrating factor method, so what did you try?
• Mar 17th 2010, 01:26 PM
jamm89
I have got this far but not sure what to do next.
$\frac{d\theta}{dt}$e^0.1t = 5e^0.1t - 2.5e^0.1t
Thanks
• Mar 17th 2010, 01:38 PM
tom@ballooncalculus
Either you're hoping you've integrated already in which case why the differentials(?), or you're hoping you're ready to integrate, having multiplied through by the I.P - in which case, you've dropped a t on the RHS, and you didn't really grasp the method. Note the bottom rows of my pictures are the situation before integrating, which wants to be...

$\frac{d \theta}{d t} e^{\frac{1}{10} t} + \frac{1}{10} \theta\ e^{\frac{1}{10} t} = 5 e^{\frac{1}{10} t} - \frac{5}{2} t e^{\frac{1}{10} t}$

The right-hand integration will involve 'parts', for which you could assign u du and v dv as usual or, if you were feeling adventurous enough to consult my pic above, you might also want to consider the general drift as...

http://www.ballooncalculus.org/asy/maps/parts.png

But more orthodox advice will be available too...
• Mar 24th 2010, 01:38 PM
jamm89
Here is what i have got but i cant get the right answer (Crying)
Intergral (5-2.5t) e^t/10 dx

By parts
u = 5-2.5t dv/dx = e^t/10
du/dx = -2.5 v = 10e^t/10

= 10e^t/10 . 5-2.5t - intergral 10e^t/10 . -2.5
= 50e^t/10 - 25te^t/10 - intergral -25e^t/10
Intergrate again
= 50e^t/10 - 25te^t/10 - (-250e^t/10) + c

But this does not give me the answer of 60 when i make t=0 (Crying)
Can someone show me where i have gone wrong please. I have tried to used the table below but struggled.
Thanks
• Mar 24th 2010, 01:58 PM
tom@ballooncalculus
Quote:

Originally Posted by jamm89
But this does not give me the answer of 60 when i make t=0 (Crying)

It isn't meant to! Use the fact that the temperature is 60 when t=0 to determine the value of c. Then you have a complete formula for finding temperatures from t's.
• Mar 24th 2010, 02:08 PM
jamm89
Quote:

Originally Posted by tom@ballooncalculus
It isn't meant to! Use the fact that the temperature is 60 when t=0 to determine the value of c. Then you have a complete formula for finding temperatures from t's.

So i have plugged t=0 in and it gives me
60=50-0+250 +c
60=300 + c
c = -240

Then i put t=2
and i get
61.07-61.07+305.35 - 240
= 65.35
• Mar 24th 2010, 02:22 PM
tom@ballooncalculus
Quote:

Originally Posted by jamm89
So i have plugged t=0 in and it gives me
60=50-0+250 +c
60=300 + c
c = -240

Good (Clapping)

Quote:

Then i put t=2
and i get
61.07-61.07+305.35 - 240
= 65.35
(Worried) Show steps...
• Mar 25th 2010, 08:11 AM
jamm89
50e^t/10 - 25te^t/10 - (-250e^t/10) + c
I substituted t=2 and c=-240
Which gives
50e^(2/10) - ((25x2)(e^2/10))- (-250xe^2/10) -240
61.07 - 61.07 - (-305.35) - 240
=65.35

Is that correct?
• Mar 25th 2010, 12:07 PM
tom@ballooncalculus
You've only subbed into the RHS. Thought the LHS was theta on its own? Only when t was zero!
• Mar 29th 2010, 09:06 AM
jamm89
So what do i need to do to be able to substitute in to get theta on its own?
Thanks
• Mar 29th 2010, 09:17 AM
tom@ballooncalculus
You can sub straight in already, just on the LHS you'll have theta times a constant, and I'm sure you'll be able to deal with that - but it's a while since you showed exactly what you've arrived at, so we might be at cross purposes. Show the whole equation.
• Mar 29th 2010, 11:42 AM
jamm89
The temperature of an object at time t is found from the equation
http://www.mathhelpforum.com/math-he...0195b97e-1.gif+ 0.1θ = 5 - 2.5t
Given that when t=0, temperature is 60˚C. find the temperature when t=2 seconds

p (t) = 0.1
Q (t) = 5-2.5t
If = e^t/10
http://www.mathhelpforum.com/math-he...0195b97e-1.gife^t/10 + 0.1θe^t/10 = 5e^t/10 - 2.5te^t/10

By parts

$\int{(5-2.5t) e^ {t/10}}$
u = 5-2.5t dv/dx = $e^ {t/10}$
du/dx = -2.5 v = $10e^ {t/10}$
$\int{(5-2.5t) e^ {t/10}dx} = 10e^ {t/10}(5-2.5t) - \int{10e^{t/10}. -2.5}$

Substitute in t = 0
$=50e^{t/10} - 25te^{t/10} - \int{-25e^{t/10}}$
$=50e^{t/10} - 25te^{t/10} - (-250e^{t/10}) + c$
= 50 - 0 + 250 + c

substitute 60 in to give c
60 = 300 + c
c = - 240
Now im not sure what to do from there.
Thanks
• Mar 30th 2010, 01:48 AM
tom@ballooncalculus
All I meant by the 'whole equation' was (not the whole of your working to date but just...)

$
50\ \theta\ e^{t/10}\ =\ 50e^{t/10} - 25te^{t/10} - (-250e^{t/10}) - 240
$

(or simpler version).

Sub into that. You do see where it came from? We integrated both sides of the differential equation (after multiplying through by the I.F.) with respect to t. (In my pictures the bottom row is the differential equation.) For the RHS we needed parts, while on the left we had an exact product-rule derivative.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last