Basic Concepts — Question 3

Question 3 – Solution

Strategy. Check the functions, solve the two data equations, then use uniqueness to establish completeness.

Step 1: Verify the entire family. Differentiation gives $$y'=Ae^t-Be^{-t},\qquad y''=Ae^t+Be^{-t}=y.$$ Therefore $y''-y=0$ for every $A,B$. No root-solving method is required.

Step 2: Fit the stated data at their actual base point. Since $e^{\ln 2}=2$ and $e^{-\ln 2}=1/2$, the equations are $$2A+B/2=5/2,\qquad 2A-B/2=3/2.$$ Adding and subtracting yields $A=1$, $B=1$. Thus $$\boxed{y=e^t+e^{-t}}.$$ Its value at $\ln 2$ is $2+1/2=5/2$, and its derivative there is $2-1/2=3/2$.

Step 3: Solve the general data problem. For arbitrary $a,b$ the same equations give $$4A=a+b,\qquad B=a-b, \qquad \boxed{A=(a+b)/4,\quad B=a-b}.$$ There is exactly one coefficient pair for each $(a,b)\in\mathbb R^2$. Substitution returns $y(\ln 2)=a$ and $y'(\ln 2)=b$, so the claim includes all data, not only the numerical example.

Step 4: Establish completeness, not just verification. Here $p=0$, $q=-1$, $g=0$ are continuous on $\mathbb R$. Any solution has some data $a,b$ at $\ln 2$. Step 3 constructs a member of the displayed family with those same data; uniqueness forces agreement on $\mathbb R$.

By contrast, $Ce^t+De^t=(C+D)e^t$ has only one effective parameter. Every such function satisfies $y'=y$, so at the base point it can realize only data with $b=a$. Two symbols for constants do not necessarily provide two independent freedoms.