# Simple homogenous 2nd order Diff EQ, but i'm stuck on the process!

1. Feb 8, 2006

### mr_coffee

Hello everyone, we just started 2nd order differentials, and i was loooking at his example and it made senes but now i'm doing the hw and i'm stuck.

Here is the problem:
Find y as a function of t if
y'' - 3y' = 0,
y(0) = 9, y(1) = 7 .
y(t) =?

Well there is my work
http://img135.imageshack.us/img135/495/lastscan6vi.jpg [Broken]

Last edited by a moderator: May 2, 2017
2. Feb 8, 2006

### benorin

you have it right, just need to solve for c1 and c2 from this system (same as you have):

$$c_1+c_2=9$$
$$c_1+c_2e^3=7$$

3. Feb 8, 2006

### benorin

Use matrix methods: the system is

$$\left(\begin{array}{cc}1&1\\1&e^3\end{array}\right) \left(\begin{array}{cc}c_1\\c_2\end{array}\right) =\left(\begin{array}{cc}9\\7\end{array}\right)$$

4. Feb 8, 2006

### benorin

multipling on the left by the inverse of the coefficient matrix gives

$$\left(\begin{array}{cc}c_1\\c_2\end{array}\right) = \frac{1}{(1)(e^3)-(1)(1)} \left(\begin{array}{cc}e^3&-1\\-1&1\end{array}\right) \left(\begin{array}{cc}9\\7\end{array}\right) = \frac{1}{e^3-1} \left(\begin{array}{cc}9e^3-7\\-9+7\end{array}\right) = \left(\begin{array}{cc}\frac{9e^3-7}{e^3-1}\\-\frac{2}{e^3-1}\end{array}\right)$$

so $$c_1=\frac{9e^3-7}{e^3-1}\mbox{ and }c_2=-\frac{2}{e^3-1}$$