- #1
Haths
- 33
- 0
This is a question aimed at the understanding of a step by step method for solving DEs.
consider for instance;
[tex]
\begin{equation}
\frac{d^{2} x}{d t^{2}} + 3 \frac{dt}{dt} - 4 = Ce^{itn}
\end{equation}
[/tex]
Where i is the complex notation, C some constant, and n some argument of that angle.
Now I fully understand how to solve the homogenious part of this equation (I think);
[tex]
\begin{equation}
\frac{d^{2} x}{d t^{2}} + 3 \frac{dt}{dt} - 4 = 0
\end{equation}
[/tex]
Let [tex] $ x(t) = e^{zt} $[/tex]
[tex]\therefore[/tex]
[tex] $ x'(t) = ze^{zt} $[/tex]
[tex] $ x''(t) = z^{2} e^{zt} $[/tex]
[tex]\therefore[/tex] We can write down;
[tex] $ z^{2} + 3z - 4 = 0 $[/tex]
[tex]\therefore[/tex]
z = 1
or z = -4
[tex]\therefore[/tex]
[tex]
\begin{equation}
x = Ae^{t} + Be^{-4t}
\end{equation}
[/tex]
But then I don't understand either my Lecturer or my course textbook for solving the equation for the inhomogenious case. I understand the proof for the adding together of the Particular Solution and the Complimentary Function, However how do you find a Particular Solution?
Do I do:
[tex]
\begin{equation}
x = Ae^{t} + Be^{-4t} + e^{itn}
\end{equation}
[/tex]
Or go back and write;
[tex]
\begin{equation}
z^{2} e^{zt} + 3ze^{zt} - 4e^{zt} = Ce^{itn}
\end{equation}
[/tex]
...and then what would I do?
So the question is, how do I. In a step by step method, solve for the inhomogenious case?
Cheers,
Haths
consider for instance;
[tex]
\begin{equation}
\frac{d^{2} x}{d t^{2}} + 3 \frac{dt}{dt} - 4 = Ce^{itn}
\end{equation}
[/tex]
Where i is the complex notation, C some constant, and n some argument of that angle.
Now I fully understand how to solve the homogenious part of this equation (I think);
[tex]
\begin{equation}
\frac{d^{2} x}{d t^{2}} + 3 \frac{dt}{dt} - 4 = 0
\end{equation}
[/tex]
Let [tex] $ x(t) = e^{zt} $[/tex]
[tex]\therefore[/tex]
[tex] $ x'(t) = ze^{zt} $[/tex]
[tex] $ x''(t) = z^{2} e^{zt} $[/tex]
[tex]\therefore[/tex] We can write down;
[tex] $ z^{2} + 3z - 4 = 0 $[/tex]
[tex]\therefore[/tex]
z = 1
or z = -4
[tex]\therefore[/tex]
[tex]
\begin{equation}
x = Ae^{t} + Be^{-4t}
\end{equation}
[/tex]
But then I don't understand either my Lecturer or my course textbook for solving the equation for the inhomogenious case. I understand the proof for the adding together of the Particular Solution and the Complimentary Function, However how do you find a Particular Solution?
Do I do:
[tex]
\begin{equation}
x = Ae^{t} + Be^{-4t} + e^{itn}
\end{equation}
[/tex]
Or go back and write;
[tex]
\begin{equation}
z^{2} e^{zt} + 3ze^{zt} - 4e^{zt} = Ce^{itn}
\end{equation}
[/tex]
...and then what would I do?
So the question is, how do I. In a step by step method, solve for the inhomogenious case?
Cheers,
Haths