# First order DE, using integrating factor

1. Jan 20, 2010

### James889

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?

2. Jan 20, 2010

### vela

Staff Emeritus
The independent variable in this problem is t. It's playing the role x would usually play.

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.

3. Jan 20, 2010

### James889

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?

4. Jan 20, 2010

### vela

Staff Emeritus
You can solve for y to get

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

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.

5. Jan 22, 2010

### James889

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.

6. Jan 22, 2010

### vela

Staff Emeritus
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.