Exact Equations — Question 9

Question 9 – Solution

Strategy. Use the conserved potential to test the point, then convert its residual into an error bound using a derivative estimate.

Step 1: Find the exact relation. The cross partials are both zero. A potential is $F=e^x+y+y^3/3$, and $F(0,0)=1$, so $$\boxed{e^x+y+\frac{y^3}{3}=1}.$$ The function $H(u)=u+u^3/3$ is strictly increasing, onto $\mathbb R$, and has $H'(u)=1+u^2>0$. Thus $H(y)=1-e^x$ gives exactly one differentiable value for every $x$. Differentiation recovers the original graph equation.

Step 2: Compute the residual and locate the true value. At the claimed point, $F=2-1-1/3=2/3$, so the potential residual $F-1$ is $-1/3$. The point is not on the selected level. At $x=\ln 2$, the true value satisfies $H(y_*)=-1$. Since $H(-1)=-4/3<-1<H(0)=0$, strict monotonicity gives $-1<y_*<0$. The claimed value $-1$ is too low.

Step 3: Bound the actual error. The mean value theorem gives, for some $\xi\in(-1,y_*)$, $$\frac 13=H(y_*)-H(-1)=H'(\xi)(y_*+1).$$ On $[-1,0]$, $1\le H'\le 2$, so $$\boxed{\frac 16\le y_*+1=|y_*-(-1)|\le\frac 13}.$$ The residual, together with monotonicity and derivative bounds, certifies both the direction and the size range of the error; no cubic formula is needed.