# First Order Linear Homogeneous Equation

1. Dec 10, 2008

### Nusc

First Order Linear Non-Homogeneous Equation

1. The problem statement, all variables and given/known data

I need to solve for e(t)

2. Relevant equations

Do I use Laplace Transform for the last integral?

3. The attempt at a solution
$$\begin{subequations} \begin{eqnarray} \nonumber \dot{\hat{{\cal E}}}(t) &=& -\kappa \hat{{\cal E}}(t) + \sqrt{2\kappa}\, \hat{{\cal E}}_{in}(t), \\ \end{eqnarray} \end{subequations} \begin{subequations} \begin{eqnarray} \dot{\hat{{\cal E}}} &=& -\kappa \hat{{\cal E}} + \sqrt{2\kappa}\, \hat{{\cal E}}_{in}, \\ \nonumber I &=& e^{\int \kappa dt} =ce^{\kappa t}\\ \nonumber ce^{\kappa t}\dot{\hat{{\cal E}}} + ce^{\kappa t}\kappa \hat{{\cal E}} &=& ce^{\kappa t} \sqrt{2\kappa}\, \hat{{\cal E}}_{in}, \\ \nonumber \frac{d}{dt}(\hat{{\cal E}}e^{ \kappa t}) &=& ce^{\kappa t} \sqrt{2\kappa}\, \hat{{\cal E}}_{in}\\ \nonumber \hat{{\cal E}}e^{ \kappa t} &=& \int^{\infty}_{0}ce^{\kappa t} \sqrt{2\kappa}\, \hat{{\cal E}}_{in}(t)dt\\ \nonumber \nonumber \end{eqnarray} \end{subequations}$$

Last edited: Dec 10, 2008
2. Dec 10, 2008

### Nusc

$$\dot{y}(t) = -\kappa y(t) + \sqrt{2\kappa} x_{in}(t)$$
If that notation helps.

3. Dec 10, 2008

### Nusc

The method I wrote above is wrong. From what I remember, your supposed to find the complementary solution and the particular solution yc, yp.

For the yp, your supposed to choose a form so as to solve for yp but x_{in} is some arbitrary function. What do I do?

4. Dec 11, 2008

### HallsofIvy

Staff Emeritus
It would help a lot if you didn't assume people will understand what problem you are doing. Is $\epsilon_{in}$ a given function and not dependent on $\epsilon$?

Assuming that it is, you are essentially using "variation of parameters".
The homogeneous equation is $\epsilon'= -\kappa\epsilon$ which should be easy to solve.

Now assume a solution of the form $\epsilon(t)= u(t)Y(t)$ where "Y(t)" is the solution to the homogeneous equation. The $\epsilon'= u'Y+ uY'$ and so the equation becomes $u'Y+ uY'= u(-\kappa Y)+ \sqrt{2\kappa}\epsilon_{in}$.

Since Y is a solution to the homogeneous equation, $Y'= -\kappa Y$, $uY'= u(-\kappa Y)$ so we can cancel those and have $u' Y= \sqrt{2\kappa}\epsilon_{in}$ remaining. Since both Y and $\epsilon_{in}$ are known functions, that's just an integration,
$$u(t)= \int_0^t \frac{\sqrt{2\kappa}\epsilon_{in}(\tau)}{Y(\tau)} d\tau[/itex] That is almost what you have. But note that the integral is NOT a definite integral from 0 to infinity. It is from 0 to t (actually, you could use any lower limit- that just changes the constant of integration) and notice that I have changed the "dummy" variable inside the integral to $\tau$ so as not to confuse it with t. Instead of what you have you should have [tex]e^{-\kappa t}u(t)= e^{\kappa t}\int_0^t e^{\kappa\tau}\epsilon_{in}(\tau)dt$$
is a specific solution, to be added to $Ce^{-\kapa t}$, the general solution to the homogeneous equation. Note that changing the variable inside the integral makes it clear that you cannot cancel the two exponentials!

Last edited: Dec 11, 2008
5. Dec 11, 2008

### Nusc

Where is the sqrt(2k)? I think my approach is correct except the limits on the integral.
I didn't realize I was doing variation of parameters.

Then how do you solve that last integraL?

$$\epsilon(t)= e^{-\kappa t}\int_0^t \sqrt{2\kappa}e^{\kappa\tau}\epsilon_{in}(\tau)dt = e^{-\kappa t} \sqrt{2\kappa}e^{\kappa\tau}\epsilon_{in}(\tau)t$$

?

Last edited: Dec 11, 2008
6. Dec 12, 2008

### Nusc

Or is that not right?

7. Dec 13, 2008

### Nusc

Why did you use tau?