Linear ODE Question (Repeated roots)

Question

$$ \displaystyle \begin{cases} \displaystyle \frac {dA}{dt} = 2A - B \\ \displaystyle \frac {dB}{dt} = A \end{cases} $$

With Initial Conditions, $$ \begin{cases} A(0) = A_0 \\ B(0) = B_0 \end{cases} $$

I differentiated $dA/dt$ and substituted $dB/dt$ into $d^2A/dt^2$ and found the roots to be repeated roots, $r_1=r_2=1$ leading to a general solution of $$A=e^t(C1 + C2)$$

How do I solve further?

Answer 1

You have that $$A''-2A'+A=0 \implies r^2-2r+1=0 \implies r=1$$ Then you have the solution $$A(t)=c_1e^t+c_2te^t$$ And for $B(t)$ $$B(t)=c_3e^t+c_4te^t$$ Multiply by t the second solution ...

Then apply the initial conditions to find $c_1,c_2$

I let you finish, I am sure you can solve the system...

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Edit

You should end with this result if I made no mistake

$$ \begin{cases} A(t)=A_0e^t+(A_0-B_0)te^t \\ B(t)=B_0e^t+(A_0-B_0)te^t \end{cases} $$

Answer 2

Your $$A=e^t(C1 + C2)$$ should have been $$A=e^t(C_1 + C_2t)$$

You can find $B$ from $$ B=2A-A'$$ and find constants from initial values.