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First order DE, using integrating factor

by James889
Tags: factor, integrating, order
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James889
#1
Jan20-10, 03:38 PM
P: 184
Hi,

I tried to solve this by using the integrating factor technique
[tex]\begin{cases}
dy/dt +10y = 1 \\
y(1/10) = 2/10
\end{cases}[/tex]

So [tex]p(x) = 10t \rightarrow e^{10t}[/tex]

[tex]e^{10t} \cdot \frac{dy}{dt} + e^{10t} \cdot 10y = e^{10t}[/tex]

This part is confusing to me, i have two different variables y and t.

How do i integrate the left side?
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vela
#2
Jan20-10, 03:45 PM
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Quote Quote by James889 View Post
[tex]e^{10t} \cdot \frac{dy}{dt} + e^{10t} \cdot 10y = e^{10t}[/tex]

This part is confusing to me, i have two different variables y and t.
The independent variable in this problem is t. It's playing the role x would usually play.

How do i integrate the left side?
The point of the integrating factor is to turn the lefthand side into a derivative. In your case, you have

[tex]e^{10t} \cdot \frac{dy}{dt} + e^{10t} \cdot 10y = (e^{10t} y)'[/tex]

You can differentiate [itex]e^{10t} y[/itex] to verify that's the case. So integrating the lefthand side is trivial because you're just integrating a derivative.
James889
#3
Jan20-10, 06:10 PM
P: 184
Ok,

So after integrating both sides, i have
[tex]e^{10t}\cdot y = \frac{1}{10} e^{10t} +C[/tex]

After inserting applying the initial conditions i get:

[tex] e^{10t} \cdot \frac{1}{10} = \frac{1}{10} e^{10t} +C[/tex]

This gives that the e terms cancel and C = 2/10 but that's incorrect.

Hm, what am i missing here?

vela
#4
Jan20-10, 06:28 PM
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First order DE, using integrating factor

Quote Quote by James889 View Post
Ok,

So after integrating both sides, i have
[tex]e^{10t}\cdot y = \frac{1}{10} e^{10t} +C[/tex]
You can solve for y to get

[tex]y=\frac{1}{10}+C e^{-10t}[/tex]

After inserting applying the initial conditions i get:

[tex] e^{10t} \cdot \frac{1}{10} = \frac{1}{10} e^{10t} +C[/tex]

This gives that the e terms cancel and C = 2/10 but that's incorrect.

Hm, what am i missing here?
A few mistakes in what you did: you didn't substitute a value in for t, and you plugged in the wrong initial value for y. Also, you apparently did the algebra wrong because the exponentials can't cancel.
James889
#5
Jan22-10, 04:19 PM
P: 184
Quote Quote by vela View Post
You can solve for y to get

[tex]y=\frac{1}{10}+C e^{-10t}[/tex]
Hm, after plugging in the right value for y i get, like you.

[tex]e^1\cdot \frac{2}{10} = \frac{e^1}{10} + C[/tex] which clearly gives C = e/10

But how did you get [tex]e^{-10t}[/tex] ?

That must be from dividing by 2e^10 but why does the variable t reappear when you already plugged in 1/10 ?

Sorry for the ignorance, i suck.
vela
#6
Jan22-10, 10:26 PM
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I divided both sides of the equation by [itex]e^{10t}[/itex] to solve for y. All you have to do now is take the value of C you found and plug it back into the solution.


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