coupled differential equation

**cooltowns** · May 24th 2010, 02:29 AM

Hi my question is

use any appropriate method to find the general solution to the coupled system of ODEs.
$<br /> \dot{y}-y+z=e^{-t}$

$\dot{z}-2z=3e^{-t}$

i tried to differentiate first equation. but wasn't getting anywhere.

**Chris L T521** · May 24th 2010, 03:09 AM

Originally Posted by cooltowns

Hi my question is

use any appropriate method to find the general solution to the coupled system of ODEs.
$<br /> \dot{y}-y+z=e^{-t}$

$\dot{z}-2z=3e^{-t}$

i tried to differentiate first equation. but wasn't getting anywhere.

You can first solve the second equation for z(t) using the integrating factor technique, and then substitute your solution into the first equation.

If you solve the second equation, you should get $z(t)=-e^{-t}+Ce^{2t}$ (verify). So the first equation becomes $\dot{y}-y=e^{-t}-(-e^{-t}+Ce^{2t})\implies \dot{y}-y=2e^{-t}+Ce^{2t}$

I leave it to you to finish this problem.

**GeoC** · May 24th 2010, 03:17 AM

Solve the homogeneous system of equations first (i.e. set the rhs to zero), and use the obtained solutions as the eigenfunctions of the system for the general solution. Make sense?

**HallsofIvy** · May 24th 2010, 03:41 AM

Yes, but since this system is "partially uncoupled" that is the hard way. As Chris L T521 said, solve the second equation for z, then put that into the first equation.

**cooltowns** · May 24th 2010, 03:55 AM

Is this possible to do it this way ?

$z=e^{-t}-\dot{y}+y$ hence $\dot{z}=-e^{-t}-\ddot{y}+\dot{y}$ Using this information and substituting it into second equation

we get $(-e^{-t}-\ddot{y}+\dot{y})-2(e^{-t}-\dot{y}+y)=3e^{-t}$
$<br /> -\ddot{y}+3\dot{y}-2y=0$

which gives us the polynomial $-\lambda^{2}+3\lambda-2=0$ which solves to $(\lambda-1)(\lambda-2)$ which gives us $y=Ae^{t}+Be^{2t}$

and then i can jst solve for z using intergrating factor method.

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