Repeated Eigenvalues — Question 3

Question 3 – Solution

Strategy. The generalized vector cancels the extra derivative introduced by the polynomial factor.

Step 1: Build a chain explicitly. The characteristic polynomial is $(\lambda+2)^2$. Let $N=A+2I=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}$; then $N^2=0$. Its kernel is $\operatorname{span}(1,1)^T$. With $v=(1,1)^T$, choose $w=(1,0)^T$, since $Nw=v$. The two vectors are independent.

Step 2: Verify the missing solution. Set $U=e^{-2t}v$ and $V=e^{-2t}(w+tv)$. Using $Nv=0$, $Nw=v$, differentiation gives $U'=AU$ and $V'=AV$. Their determinant is $e^{-4t}\det(v,w)=-e^{-4t}\ne 0$, so they form a complete real basis. For $Z=te^{-2t}v$, however, $Z'-AZ=e^{-2t}v\ne 0$: the polynomial term by itself is not a solution.

Step 3: Determine the IVP coefficients. Since $(2,1)^T=v+w$, the required solution is $U+V$: $$\boxed{X=e^{-2t}\binom{2+t}{1+t}.}$$ At zero it is $(2,1)^T$, and its derivative is $(-3,-1)^T$, agreeing with $A(2,1)^T$. The mode calculation verifies the equation for every time, not just at the initial point.

Step 4: Account for chain freedom. The equation $Nw=v$ means $w_1-w_2=1$, so every choice is $w_c=(1+c,c)^T=w+cv$, $c\in\mathbb R$. The new second solution is $V_c=V+cU$. Hence the same IVP becomes $\boxed{X=(1-c)U+V_c}$. In general a combination $aU+bV$ becomes $(a-bc)U+bV_c$. The basis changes, but its span and every physical solution remain the same. Scaling or shifting a chain must be accompanied by the corresponding coefficient change.