# First order DE, using integrating factor

by James889
Tags: factor, integrating, order
 P: 184 Hi, I tried to solve this by using the integrating factor technique $$\begin{cases} dy/dt +10y = 1 \\ y(1/10) = 2/10 \end{cases}$$ So $$p(x) = 10t \rightarrow e^{10t}$$ $$e^{10t} \cdot \frac{dy}{dt} + e^{10t} \cdot 10y = e^{10t}$$ 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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PF Gold
P: 11,755
 Quote by James889 $$e^{10t} \cdot \frac{dy}{dt} + e^{10t} \cdot 10y = e^{10t}$$ 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

$$e^{10t} \cdot \frac{dy}{dt} + e^{10t} \cdot 10y = (e^{10t} y)'$$

You can differentiate $e^{10t} y$ to verify that's the case. So integrating the lefthand side is trivial because you're just integrating a derivative.
 P: 184 Ok, So after integrating both sides, i have $$e^{10t}\cdot y = \frac{1}{10} e^{10t} +C$$ After inserting applying the initial conditions i get: $$e^{10t} \cdot \frac{1}{10} = \frac{1}{10} e^{10t} +C$$ This gives that the e terms cancel and C = 2/10 but that's incorrect. Hm, what am i missing here?
Emeritus
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PF Gold
P: 11,755
First order DE, using integrating factor

 Quote by James889 Ok, So after integrating both sides, i have $$e^{10t}\cdot y = \frac{1}{10} e^{10t} +C$$
You can solve for y to get

$$y=\frac{1}{10}+C e^{-10t}$$

 After inserting applying the initial conditions i get: $$e^{10t} \cdot \frac{1}{10} = \frac{1}{10} e^{10t} +C$$ 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.
P: 184
 Quote by vela You can solve for y to get $$y=\frac{1}{10}+C e^{-10t}$$
Hm, after plugging in the right value for y i get, like you.

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

But how did you get $$e^{-10t}$$ ?

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.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,755 I divided both sides of the equation by $e^{10t}$ 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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