Laplace Transforms — Question 1

Question 1 – Solution

Strategy. Keep the initial data on the right side of the transformed system, then distinguish component responses from observed combinations.

Step 1: Solve the algebraic system. The derivative rule gives $(s+2)F-G=1$ and $-2F+(s+3)G=0$. Their determinant is $(s+1)(s+4)$, so $$F=\frac{s+3}{(s+1)(s+4)},\qquad G=\frac{2}{(s+1)(s+4)}.$$ Both transforms converge for $\operatorname{Re}s>-1$. Dropping the initial term $1$ would incorrectly give the zero solution.

Step 2: Invert and check. Partial fractions give $$\boxed{x=\tfrac 23e^{-t}+\tfrac 13e^{-4t},\qquad y=\tfrac 23(e^{-t}-e^{-4t}).}$$ At zero these are $(1,0)$. Differentiation gives $x'=-\tfrac 23e^{-t}-\tfrac 43e^{-4t}$ and $y'=-\tfrac 23e^{-t}+\tfrac 83e^{-4t}$, which equal $-2x+y$ and $2x-3y$.

Step 3: Explain the cancelled pole. Subtracting the transforms gives $F-G=1/(s+4)$, hence $\boxed{w=e^{-4t}}$. The slow mode is a multiple of $(1,1)^T$; subtracting its components annihilates it. The state still contains this mode. Cancellation in an observation does not remove a mode from the system.

Step 4: Locate the component peak. Since $y'=\tfrac 23e^{-t}(4e^{-3t}-1)$, it changes sign once, from positive to negative, at $t_*=\log(4)/3$. Thus $$\boxed{\max_{t\ge 0}y(t)=\tfrac 12\,4^{-1/3},\qquad t_*=\log(4)/3.}$$ Here $y'(0)=2$ because the first component feeds the second. Decay of each mode does not imply monotonic decay of every component: their signed combination initially grows. The curve tends back to zero.