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First order DE, using integrating factor 
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#1
Jan2010, 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? 


#2
Jan2010, 03:45 PM

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[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. 


#3
Jan2010, 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? 


#4
Jan2010, 06:28 PM

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First order DE, using integrating factor
[tex]y=\frac{1}{10}+C e^{10t}[/tex] 


#5
Jan2210, 04:19 PM

P: 184

[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. 


#6
Jan2210, 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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