# Solutions to differential equations

• Nov 15th 2008, 12:12 PM
Scopur
Solutions to differential equations
I have (okay i cant get it to work.. it shows up in my latex but not here.. so here it is in non- latex format)

d/dt $[y_1 y_2 y_3]^T$ = \left[ \begin {array}{ccc} 1&0&0\\\noalign{\medskip}0&-1&1 \\\noalign{\medskip}0&0&-1\end {array} \right] $[y_1 y_2 y_3]^T$

I need to solve explicitly for y .. but i dont know how to sovle for $y_2(t)$ explicitly.

-frig hold on im not sure wat my syntax error is.
• Nov 15th 2008, 12:14 PM
Math_Helper
Post the equation that you need resolved, so I can see it.
• Nov 15th 2008, 12:18 PM
Scopur
okay i got it up but its a lil ugly looking :P
• Nov 15th 2008, 12:38 PM
Math_Helper
I don't now ho to do (Worried)
Sorry
• Nov 15th 2008, 01:35 PM
HallsofIvy
Quote:

Originally Posted by Scopur
I have (okay i cant get it to work.. it shows up in my latex but not here.. so here it is in non- latex format)

d/dt $[y_1 y_2 y_3]^T$ = \left[ \begin {array}{ccc} 1&0&0\\\noalign{\medskip}0&-1&1 \\\noalign{\medskip}0&0&-1\end {array} \right] $[y_1 y_2 y_3]^T$

I need to solve explicitly for y .. but i dont know how to sovle for $y_2(t)$ explicitly.

-frig hold on im not sure wat my syntax error is.

This is about as close to trivial as you can get! That matrix equation reduces very quickly to three equations:
$\frac{dy_1}{dt}= y_1$
$\frac{dy_2}{dt}= -y_2+ y_3$
$\frac{dy_3}{dt}= -y_3$

You should be able to solve for $y_1$ and $y_3$ and apparently you have done that since you are only asking about $y_2$. Now put the function you got for $y_3$ into the equation for $y_2$ and you have a non-homogeneous equation for $y_2[/sup]. Are you saying you cannot solve that? I presume you got $y_1(t)= Ce^t/$ and $y_2(t)= De^{-t}$. Putting that into the second equation, [tex]dy_2/dt= -y_2+ De^{-t}$. That's a linear equation with "integrating factor" [tex]e^{t}[/mathMultiplying the entire equation by $e^t$, we have $e^t dy_2/dt+ e^ty_2= d(e^ty_2)/dt= D$ so $e^ty= Dt+ E$ and $y_2(t)= Dte^{-t}+ Ee^{-t}$