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

[tex]
\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}
[/tex]

 

[Nusc]

[tex]
\dot{y}(t) = -\kappa y(t) + \sqrt{2\kappa} x_{in}(t)
[/tex]
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 [itex]\epsilon_{in}[/itex] a given function and not dependent on [itex]\epsilon[/itex]?

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

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

Since Y is a solution to the homogeneous equation, [itex]Y'= -\kappa Y[/itex], [itex]uY'= u(-\kappa Y)[/itex] so we can cancel those and have [itex]u' Y= \sqrt{2\kappa}\epsilon_{in}[/itex] remaining. Since both Y and [itex]\epsilon_{in}[/itex] 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]

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 [itex]\tau[/itex] 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[/tex]
is a specific solution, to be added to [itex]Ce^{-\kapa t}[/itex], 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!

 

[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?

[tex]
\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
[/tex]

?

 

[Nusc]

Or is that not right?

 

[Nusc]

Why did you use tau?