Basic Concepts — Question 2

Question 2 – Solution

Strategy. Integrate each derivative level separately, then examine which independent data the constants control.

Step 1: Recover the full family. One integration gives $y'=3t^2+A$; another gives $$\boxed{y=t^3+At+B},\qquad A,B\in\mathbb R.$$ These are all solutions on any interval: if two functions have second derivative $6t$, their difference has zero second derivative and therefore is affine. The constant $A$ changes velocity; $B$ changes position without changing velocity.

Step 2: Fit and verify both data. At $t=1$, the conditions become $1+A+B=2$ and $3+A=-1$. Thus $A=-4$, $B=5$, giving $$\boxed{y=t^3-4t+5}.$$ Indeed, $y'=3t^2-4$, $y''=6t$, $y(1)=2$, and $y'(1)=-1$. The polynomial is defined on all of $\mathbb R$.

Step 3: Locate the missing freedom. Every function $t^3+C$ solves the differential equation, but every one has derivative $3t^2$. In particular, its slope at $t=1$ is always 3. It cannot realize the required slope $-1$, regardless of $C$. Verification of a family does not by itself prove that the family is general.

Step 4: Separate position from velocity. With only $y(1)=2$, we obtain $B=1-A$, so all possibilities are $$\boxed{y=t^3+A(t-1)+1},\qquad y'(1)=3+A.$$ The initial slope ranges over every real number as $A$ varies. Setting it to $-1$ forces $A=-4$ and hence the single solution above.

Alternatively, the difference $w$ of any two solutions with both prescribed data satisfies $w''=0$, $w(1)=w'(1)=0$. Writing $w=at+b$ gives $a=0$ and $b=0$. This proves uniqueness directly, rather than merely counting constants.