Introduction: In this question, we explain how to solve system of initial value problems example is given in the following solution.
Question: Solve system of initial value problems
\[u^\prime=-0.2\left(v-2\right), u\left(0\right)=2 \]
\[v^\prime=0.8\left(u-2\right), v\left(0\right)=0 \]
Solution:
Given system of IVPs
\[u^\prime=-0.2\left(v-2\right), u\left(0\right)=2 \] (1)
\[v^\prime=0.8\left(u-2\right), v\left(0\right)=0 \] (2)
If we differentiate eqn. (1)
\[u^{\prime\prime}=-0.2v\prime\]
and Use eqn. (2) then
\[u^{\prime\prime}=-0.2\left(0.8\left(u-2\right)\right)\]
or
\[u^{\prime\prime}=-0.16u+0.32\]
or
\[u^{\prime\prime}+0.16u-0.32=0\]
Now this is a second order linear ode .
Writing it’s Auxiliary equation
\[m^2+0.16m-0.4=0\]
\[m=-0.651, 0.491\]
are roots.
Now general solution can be written as
\[u\left(t\right)=c_1e^{-0.651t}+c_2e^{0.491t}\]
Using initial condition for u
\[u\left(0\right)=2\]
we get
\[2=c_1+c_2\]
And from eqn (1)
\[u^\prime\left(0\right)=-0.2\left(v\left(0\right)-2\right)=-0.2\left(0-2 \right)=0.4\]
and also
\[ u^\prime\left(t\right)=-0.651c_1e^{-0.651t}+0.491c_2e^{0.491t}\]
\[u^\prime\left(0\right)=0.4=-0.651c_1+0.491c_2\]
hence we solve these and get
\[ c_1=0.51, c_2=1.49\]
now u(t) is
\[u\left(t\right)=0.51e^{-0.651t}+1.49e^{0.491t}\]
Now using eqn. (1)
\[u^\prime=-0.2\left(v-2\right)\]
\[-0.51\ast0.651e^{-0.651t}+1.49\ast0.491e^{0.491t}=-0.2\left(v-2\right)\]
this gives v(t)
\[v\left(t\right)=2+1.66e^{-0.651t}-3.65e^{0.491t}\]
Thus u(t) , v(t) are solutions of system of IVPs.