First Order Linear Homogeneous Equation

[Nusc]

First Order Linear Non-Homogeneous Equation

Homework Statement

I need to solve for e(t)

Homework Equations

Do I use Laplace Transform for the last integral?

The Attempt at a Solution

$$\begin{subequations}<br /> \begin{eqnarray}<br /> \nonumber<br /> \dot{\hat{{\cal E}}}(t) &=& -\kappa \hat{{\cal E}}(t) + \sqrt{2\kappa}\, \hat{{\cal E}}_{in}(t), \\ <br /> <br /> \end{eqnarray}<br /> \end{subequations}<br /> <br /> <br /> <br /> <br /> \begin{subequations}<br /> \begin{eqnarray}<br /> \dot{\hat{{\cal E}}} &=& -\kappa \hat{{\cal E}} + \sqrt{2\kappa}\, \hat{{\cal E}}_{in}, \\ <br /> \nonumber<br /> I &=& e^{\int \kappa dt} =ce^{\kappa t}\\ <br /> \nonumber<br /> ce^{\kappa t}\dot{\hat{{\cal E}}} + ce^{\kappa t}\kappa \hat{{\cal E}} &=& ce^{\kappa t} \sqrt{2\kappa}\, \hat{{\cal E}}_{in}, \\ <br /> \nonumber<br /> \frac{d}{dt}(\hat{{\cal E}}e^{ \kappa t}) &=& ce^{\kappa t} \sqrt{2\kappa}\, \hat{{\cal E}}_{in}\\<br /> \nonumber <br /> \hat{{\cal E}}e^{ \kappa t} &=& \int^{\infty}_{0}ce^{\kappa t} \sqrt{2\kappa}\, \hat{{\cal E}}_{in}(t)dt\\<br /> \nonumber<br /> \nonumber<br /> \end{eqnarray}<br /> \end{subequations}$$

 

[Nusc]

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

 

[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?

 

[HallsofIvy]

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,
[tex]u(t)= \int_0^t \frac{\sqrt{2\kappa}\epsilon_{in}(\tau)}{Y(\tau)} d\tau[/itex]<br /> <br /> That is <b>almost</b> 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.<br /> <br /> Instead of what you have you should have<br /> $$e^{-\kappa t}u(t)= e^{\kappa t}\int_0^t e^{\kappa\tau}\epsilon_{in}(\tau)dt$$ <br /> 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 <b>cannot</b> cancel the two exponentials![/tex]

 

[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?

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

?

 

[Nusc]

Or is that not right?

 

[Nusc]

Why did you use tau?