# Find Particular and then General Solution of Complex Function

• Dec 11th 2010, 11:25 AM
Stylis10
Find Particular and then General Solution of Complex Function
I dont quite know how to begin with this question.

$\displaystyle \dot{z} = \lambda z + g(t)$ (1)
where $\displaystyle \lambda \in \mathbb{C}$, and $\displaystyle g(t)$ is a continuous complex function of real variable.
a) Derive the variation of constants formula for finding a particular solution of (1).
b) Using this formula, find the general solution of (1) with $\displaystyle g(t) = cos(t) + i sin(2t).$

I tried to write z as (x+iy) and find a solution using variable of constants(parameters) method, but have come to no avail, i missed the lecture which covered this but even with borrowing someone elses notes i cant seem to work it out, neither can most people ive spoken to, how would i go about doing this question?
• Dec 11th 2010, 04:53 PM
Prove It
When you are solving this DE, I assume you are solving for $\displaystyle \displaystyle z$...

$\displaystyle \displaystyle \frac{dz}{dt} = \lambda z + g(t)$

$\displaystyle \displaystyle \frac{dz}{dt} - \lambda z = g(t)$

This is first order linear, so the Integrating Factor is $\displaystyle \displaystyle e^{\int{-\lambda \,dt}} = e^{-\lambda t}$.

Multiplying both sides by the Integrating Factor gives

$\displaystyle \displaystyle e^{-\lambda t}\,\frac{dz}{dt} - \lambda e^{-\lambda t}z = e^{-\lambda t}g(t)$

$\displaystyle \displaystyle \frac{d}{dt}(e^{-\lambda t}z) = e^{-\lambda t}g(t)$

$\displaystyle \displaystyle e^{-\lambda t}z = \int{e^{-\lambda t}g(t)\,dt}$

$\displaystyle \displaystyle z = e^{\lambda t}\int{e^{-\lambda t}g(t)\,dt}$.
• Dec 12th 2010, 01:49 AM
Stylis10
Quote:

Originally Posted by Prove It
When you are solving this DE, I assume you are solving for $\displaystyle \displaystyle z$...

$\displaystyle \displaystyle \frac{dz}{dt} = \lambda z + g(t)$

$\displaystyle \displaystyle \frac{dz}{dt} - \lambda z = g(t)$

This is first order linear, so the Integrating Factor is $\displaystyle \displaystyle e^{\int{-\lambda \,dt}} = e^{-\lambda t}$.

Multiplying both sides by the Integrating Factor gives

$\displaystyle \displaystyle e^{-\lambda t}\,\frac{dz}{dt} - \lambda e^{-\lambda t}z = e^{-\lambda t}g(t)$

$\displaystyle \displaystyle \frac{d}{dt}(e^{-\lambda t}z) = e^{-\lambda t}g(t)$

$\displaystyle \displaystyle e^{-\lambda t}z = \int{e^{-\lambda t}g(t)\,dt}$

$\displaystyle \displaystyle z = e^{\lambda t}\int{e^{-\lambda t}g(t)\,dt}$.

I had never assumed $\displaystyle \dot{z}$ to be the derivative of z w.r.t t, i though it was the complex conjugate of z. the question for the amount of marks given for it seems too simple for it to be the derivative. Is there anyway it could be the complex conjugate instead? Or is that just not possible.
• Dec 12th 2010, 01:59 AM
Prove It
Quote:

Originally Posted by Stylis10
I had never assumed $\displaystyle \dot{z}$ to be the derivative of z w.r.t t, i though it was the complex conjugate of z. the question for the amount of marks given for it seems too simple for it to be the derivative. Is there anyway it could be the complex conjugate instead? Or is that just not possible.

A dot above a variable nearly always means that derivative with respect to time... If it was a conjugate it should be a line...
• Dec 12th 2010, 06:13 AM
Jester
Quote:

Originally Posted by Prove It
When you are solving this DE, I assume you are solving for $\displaystyle \displaystyle z$...

$\displaystyle \displaystyle \frac{dz}{dt} = \lambda z + g(t)$

$\displaystyle \displaystyle \frac{dz}{dt} - \lambda z = g(t)$

This is first order linear, so the Integrating Factor is $\displaystyle \displaystyle e^{\int{-\lambda \,dt}} = e^{-\lambda t}$.

Multiplying both sides by the Integrating Factor gives

$\displaystyle \displaystyle e^{-\lambda t}\,\frac{dz}{dt} - \lambda e^{-\lambda t}z = e^{-\lambda t}g(t)$

$\displaystyle \displaystyle \frac{d}{dt}(e^{-\lambda t}z) = e^{-\lambda t}g(t)$

$\displaystyle \displaystyle e^{-\lambda t}z = \int{e^{-\lambda t}g(t)\,dt}$

$\displaystyle \displaystyle z = e^{\lambda t}\int{e^{-\lambda t}g(t)\,dt}$.

$\displaystyle + c$
• Dec 12th 2010, 03:36 PM
Prove It
Quote:

Originally Posted by Danny
$\displaystyle + c$

That would be implied by the fact that there is an integral that can only be solved once we know what $\displaystyle \displaystyle g(t)$ is.
• Dec 12th 2010, 04:10 PM
Jester
I disagree. You're being sloppy.