SOLUTION: First-order sysytem ff linear differential equations
1. x'=-x, y' =x-y, z' =y-z, x(0) = 2, y(0) = 1, z(0)= 0
2. x'=y-z, y' =z-x, z' =x-y, x(0) = 2, y(0) = -1, z(0)= -1
Algebra.Com
Question 1160294: First-order sysytem ff linear differential equations
1. x'=-x, y' =x-y, z' =y-z, x(0) = 2, y(0) = 1, z(0)= 0
2. x'=y-z, y' =z-x, z' =x-y, x(0) = 2, y(0) = -1, z(0)= -1
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ =\ -x(t);\ x(0)\ =\ 2)
\ =\ x(t)\ -\ y(t);\ y(0)\ =\ 1)
\ =\ y(t)\ -\ z(t);\ z(0)\ =\ 0)
=========================================
)}{dt}\ =\ -x(t))
)}{dt}\ +\ x(t)\ =\ 0)
Let \ =\ e^{\small{\int\,1\,dt}}\ =\ e^t)
and multiply by )
)}{dt}\ +\ e^tx(t)\ =\ 0)
Substitute 
)}{dt}\ +\ \frac{d}{dt}e^tx(t)\ =\ 0)
Reverse the product rule: )
\)\ =\ 0)
Integrate
\ =\ C)
where 
is an arbitrary constant
Divide both sides by
\ =\ Ce^{-t})
\ =\ 2\ \Right\ C\ =\ 2)
\ =\ 2e^{-t})
====================================================
\ =\ x(t)\ -\ y(t))
\ +\ y(t)\ =\ 2e^{-t})
)}{dt}\ =\ 2e^{-t}\ -\ y(t))
Let and multiply by
)}{dt}\ +\ e^ty(t)\ =\ 2)
Substitute
)}{dt}\ +\ \frac{d}{dt}e^ty(t)\ =\ 2)
Reverse the product rule:
\)\ =\ 2)
Integrate
\ =\ 2t\ +\ C)
where is an arbitrary constant
Divide both sides by
\ =\ e^{-t}(2t\ +\ C))
\ =\ 1\ \Right\ C\ =\ 1)
\ =\ e^{-t}(2t\ +\ 1))
=====================================
Same process for )
which yields
\ =\ e^{-t}t(t\ +\ 1))
John
My calculator said it, I believe it, that settles it
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