Linear Equations — Question 8

Question 8 – Solution

Strategy. Integrate separately on the two smooth pieces and match values at the switch; continuity need not imply a continuous derivative.

Step 1: Solve and match. Before the switch, $(e^ty)'=2e^t$, so $y=2+Ce^{-t}$. Initial data give $C=-2$ and $y(1)=A:=2(1-e^{-1})$. Afterward $(e^ty)'=0$; matching this value yields $$\boxed{y(t)=\begin{cases} 2(1-e^{-t}),&0\le t\le 1,\\ A e^{-(t-1)},&t>1, \end{cases}\qquad A=2(1-e^{-1}).}$$ Differentiation on each open piece verifies its corresponding equation.

Step 2: Maximum and derivative jump. Before $1$, $y'=2e^{-t}>0$; after $1$, $y'=-Ae^{-(t-1)}<0$. Hence the global maximum is $\boxed{A\text{ at }t=1}$, and the limit is zero. At the switch, $$y'(1^-)=2/e,\qquad y'(1^+)=-A=2/e-2.$$ The jump is $-2$, so the derivative does not exist there. The response meets the requested piecewise conditions, but is not a classical differentiable solution at the switch.

Step 3: Total response. Direct integration gives $$\int_0^1 2(1-e^{-t})dt=2/e,\qquad \int_1^\infty Ae^{-(t-1)}dt=A,$$ so $\boxed{\int_0^\infty y(t)dt=2=\int_0^\infty f(t)dt}$. Equivalently, integrate $y'+y=f$ on the two pieces: continuity cancels the joining values and $y(0)=\lim y=0$ cancels the endpoint contribution.