Definitions — Question 2

Question 2 – Solution

Strategy. Verify two separate requirements: the differential equation must hold throughout an interval, and the initial value must hold at the specified point.

Step 1: Residuals. Direct differentiation gives $$\begin{aligned} f'&=4e^{-2x}, &R[f]&=4e^{-2x}+6-4e^{-2x}-6=0,\\ g'&=-4e^{2x}, &R[g]&=-4e^{2x}+6-4e^{2x}-6=-8e^{2x},\\ h'&=0, &R[h]&=0+6-6=0. \end{aligned}$$ Also $f(0)=1$, $g(0)=1$, and $h(0)=3$.

Step 2: Decisions. Since $e^{2x}>0$, the residual for $g$ never vanishes. Hence $$\boxed{\begin{gathered} f\text{ and }h\text{ solve the differential equation on }\mathbb R,\\ f\text{ alone among these candidates solves the IVP on }\mathbb R. \end{gathered}}$$ The function $g$ passes the initial-value test but fails the equation; $h$ does the reverse.

Step 3: Parameter selection. For $y=3+Ce^{-2x}$, $$y'=-2Ce^{-2x},\qquad y'+2y=6.$$ Every real $C$ gives an equation solution on $\mathbb R$. The initial condition requires $3+C=1$, so $\boxed{C=-2}$, recovering $f$.

Step 4: Logical distinction. An initial value is one pointwise constraint. A differential equation is an identity along the candidate function on an interval. For example, $v(x)=3+x^2$ satisfies $v'+2v=6$ at $x=0$, but its residual $2x+2x^2$ is not identically zero on any open interval. Neither pointwise test replaces interval verification.