Hi my question is
use any appropriate method to find the general solution to the coupled system of ODEs.
$\displaystyle
\dot{y}-y+z=e^{-t}$
$\displaystyle \dot{z}-2z=3e^{-t}$
i tried to differentiate first equation. but wasn't getting anywhere.
Hi my question is
use any appropriate method to find the general solution to the coupled system of ODEs.
$\displaystyle
\dot{y}-y+z=e^{-t}$
$\displaystyle \dot{z}-2z=3e^{-t}$
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 $\displaystyle z(t)=-e^{-t}+Ce^{2t}$ (verify). So the first equation becomes $\displaystyle \dot{y}-y=e^{-t}-(-e^{-t}+Ce^{2t})\implies \dot{y}-y=2e^{-t}+Ce^{2t}$
I leave it to you to finish this problem.
Is this possible to do it this way ?
$\displaystyle z=e^{-t}-\dot{y}+y$ hence $\displaystyle \dot{z}=-e^{-t}-\ddot{y}+\dot{y}$Using this information and substituting it into second equation
we get $\displaystyle (-e^{-t}-\ddot{y}+\dot{y})-2(e^{-t}-\dot{y}+y)=3e^{-t}$
$\displaystyle
-\ddot{y}+3\dot{y}-2y=0$
which gives us the polynomial$\displaystyle -\lambda^{2}+3\lambda-2=0$which solves to $\displaystyle (\lambda-1)(\lambda-2)$ which gives us $\displaystyle y=Ae^{t}+Be^{2t}$
and then i can jst solve for z using intergrating factor method.