GATE 2018 IN Q.37: solution and answer

Official answer: DSource: Official answer key, GATE 2018 IN (IIT Guwahati)

1. For the unforced system $\dot{x}=Ax$ the solution is $x(t)=e^{At}x(0)$, where $\Phi (t)=e^{At}$ is the state transition matrix.

2. Here $A=\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right]$ is diagonal, so every power of it is diagonal: $A^k=\left[\begin{array}{cc}(-1{)}^k& 0\\ 0& (-2{)}^k\end{array}\right]$. Summing the series term by term gives $$e^{At}=\sum _{k=0}^{\infty }\frac{A^k{t}^k}{k!}=\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right]$$ This is the same as saying the two states are decoupled: ${\dot{x}}_1=-x_1$ and ${\dot{x}}_2=-2x_2$.

3. Check by Laplace transform: $sI-A=\left[\begin{array}{cc}s+1& 0\\ 0& s+2\end{array}\right]$, so $(sI-A{)}^{-1}=\left[\begin{array}{cc}\frac{1}{s+1}& 0\\ 0& \frac{1}{s+2}\end{array}\right]$ and $$\Phi (t)={\mathcal{L}}^{-1}\left[(sI-A{)}^{-1}\right]=\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right]$$

4. So $x(t)=\left[\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}{e}^{-t}\\ e^{-2t}\end{array}\right]$. This is option (D).

5. Why the others are wrong. (A) and (C) have a non-zero top-right entry. Step 2 shows that $e^{At}$ of a diagonal $A$ has no off-diagonal entries at all. A term like $t{e}^{-2t}$ would need a Jordan block at $-2$, which means a repeated eigenvalue and a non-zero off-diagonal entry in $A$. This $A$ has neither. (B) has $e^{+2t}$, but the diagonal entries of $e^{At}$ are $e^{{\lambda }_{i}t}$ with ${\lambda }_1=-1$ and ${\lambda }_2=-2$, so it would need an eigenvalue at $+2$.

6. A quick check by substitution. All four matrices equal $I$ at $t=0$, so $\Phi (0)=I$ cannot tell them apart. Putting each candidate into ${\dot{x}}_1+x_1=0$ does. For (A), $x_1=e^{-t}+t{e}^{-2t}$ gives ${\dot{x}}_1+x_1=(1-t)e^{-2t}\ne 0$. For (C), $x_1=e^{-t}-t^2{e}^{-2t}$ gives ${\dot{x}}_1+x_1=t(t-2)e^{-2t}\ne 0$. For (B), $x_2=e^{2t}$ gives ${\dot{x}}_2+2x_2=4e^{2t}\ne 0$. Only (D) satisfies both state equations and the initial condition. Because the solution is unique, the answer is (D).