First Order Linear Homogeneous Equation

In summary, the conversation is about solving for e(t) in a first order linear non-homogeneous equation. The question of whether Laplace Transform should be used for the last integral is raised. The attempt at a solution involves using variation of parameters and choosing a solution of the form e(t)=u(t)Y(t). The correct method involves using a definite integral from 0 to t and adding the specific solution to the general solution of the homogeneous equation. There is a discussion about the limits and the use of the variable tau.
  • #1
Nusc
760
2
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]
 
Last edited:
Physics news on Phys.org
  • #2
[tex]
\dot{y}(t) = -\kappa y(t) + \sqrt{2\kappa} x_{in}(t)
[/tex]
If that notation helps.
 
  • #3
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
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!
 
Last edited by a moderator:
  • #5
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]

?
 
Last edited:
  • #6
Or is that not right?
 
  • #7
Why did you use tau?
 

FAQ: First Order Linear Homogeneous Equation

What is a First Order Linear Homogeneous Equation?

A First Order Linear Homogeneous Equation is a type of differential equation that can be written in the form y' + p(x)y = 0, where y' represents the derivative of y with respect to x, and p(x) is a continuous function of x. It is called "linear" because y and its derivative appear in the equation with a power of 1, and "homogeneous" because all terms involve the dependent variable y and its derivative.

What is the general solution to a First Order Linear Homogeneous Equation?

The general solution to a First Order Linear Homogeneous Equation is y = ce-∫p(x)dx, where c is an arbitrary constant and ∫p(x)dx represents the indefinite integral of p(x). This solution can be obtained using the method of separation of variables and integration.

What is the significance of the arbitrary constant in the general solution?

The arbitrary constant in the general solution represents the family of solutions to the differential equation. Each value of c will give a different specific solution, but they will all satisfy the original equation. This constant allows for a wide range of possible solutions to the equation.

What is the relationship between First Order Linear Homogeneous Equations and exponential functions?

First Order Linear Homogeneous Equations are closely related to exponential functions, as they share the same form of y = cerx, where c and r are constants. This is because the solution to a First Order Linear Homogeneous Equation involves taking the integral of a function with an exponential term.

What are some real-world applications of First Order Linear Homogeneous Equations?

First Order Linear Homogeneous Equations can be used to model a variety of natural phenomena, such as population growth, radioactive decay, and the spread of diseases. They are also commonly used in physics and engineering to describe the behavior of systems with changing variables.

Similar threads

Replies
16
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
11
Views
2K
Replies
3
Views
1K
Replies
22
Views
3K
Back
Top