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Got this problem and we've been given a program which can solve for x, for the equation:
Ax = b
Where
[tex]A = \left( \begin{array}{rrrrrr}
b & c & 0 & 0 & \cdots & 0 \\
a & b & c & 0 & \cdots & 0\\
0 & \ddots & \ddots & \ddots & & \vdots \\
\vdots & & \ddots & \ddots & \ddots & 0 \\
\vdots & & & \ddots & \ddots & c\\
0 & \cdots & \cdots & 0 & a & b \\
\end{array}
\right)[/tex]
If I left a = - 1, c = -1 and b = 2 + (1/n^2) (for my number of intervals n) and b = [1/n^2, 1/n^2, ..., 1/n^2]
Then I am approximation solutions to:
[tex]-\frac{d^2y}{dx^2} + y = 1 \quad \text{and} \quad y(0)=0 \quad y(1)=0[/tex]
I kind of understand that (or at least I have notes in front of me which seem to conclude that fairly logically). However I am now asked to set up a similar system for:
[tex]-\frac{d^2y}{dx^2} + 20 \frac{dy}{dx} = 1 \quad \text{and} \quad y(0)=0 \quad y(1)=0[/tex]
And I have no idea where to start, any help please.
Ax = b
Where
[tex]A = \left( \begin{array}{rrrrrr}
b & c & 0 & 0 & \cdots & 0 \\
a & b & c & 0 & \cdots & 0\\
0 & \ddots & \ddots & \ddots & & \vdots \\
\vdots & & \ddots & \ddots & \ddots & 0 \\
\vdots & & & \ddots & \ddots & c\\
0 & \cdots & \cdots & 0 & a & b \\
\end{array}
\right)[/tex]
If I left a = - 1, c = -1 and b = 2 + (1/n^2) (for my number of intervals n) and b = [1/n^2, 1/n^2, ..., 1/n^2]
Then I am approximation solutions to:
[tex]-\frac{d^2y}{dx^2} + y = 1 \quad \text{and} \quad y(0)=0 \quad y(1)=0[/tex]
I kind of understand that (or at least I have notes in front of me which seem to conclude that fairly logically). However I am now asked to set up a similar system for:
[tex]-\frac{d^2y}{dx^2} + 20 \frac{dy}{dx} = 1 \quad \text{and} \quad y(0)=0 \quad y(1)=0[/tex]
And I have no idea where to start, any help please.
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