Differential

• Mar 13th 2009, 03:03 PM
ronaldo_07
Differential
http://img18.imageshack.us/img18/9839/differential.jpg
• Mar 13th 2009, 03:53 PM
HallsofIvy
$y_1'= 3y_1+ 4y_2$
$y_2'= -2y_1- y_2$

Differentiate the first equation to get $y_1"= 3y_1'+ 4y_2'$. From the second equation, $y_1"= 3y_1'+ 4(-2y_1- y_2)= 2y_1'- 8y_1- 4y_2$.

From the first equation again, $4y_2= y_1'- 3y_1$ so $y_1"= 3y_1'- 8y_1- (y_1'- 3y_1)= 2y_1'- 5y_1$.

That is, $y_1"- 2y_1'+ 5y_1= 0$. Can you solve that?

That will have, of course, two constants in the general solution. To find $y_2$, use $4y_2= y_1'- 3y_1$ so as not to introduce more constants.
• Mar 14th 2009, 11:04 AM
ronaldo_07
I got $y=Ae^{(landa-2)x}+Be^{(landa+2)x}$ is this correct?