Final Thoughts — Question 7

Question 7 – Solution

Strategy. Use interval verification for the ODE. Infinite-order agreement at one point is still pointwise information unless convergence to the function is established.

Step 1: Identical Taylor coefficients. For both functions, every derivative at zero is zero, so both Taylor series are $$\boxed{\sum_{n=0}^{\infty}\frac{0}{n!}x^n=0.}$$ This describes the series built from the derivatives; it does not yet assert that the series represents either function away from zero.

Step 2: The equation fails to the right. For $x>0$, the chain rule gives $$\boxed{v'(x)=\frac{2}{x^3}e^{-1/x^2}>0.}$$ Every open interval containing $0$ includes positive $x$, where the required identity $v'=0$ fails. Thus $v$ does not solve this IVP on such an interval, despite satisfying the initial value and having every derivative there equal to those of $u$.

Step 3: Uniqueness directly. If $y'=0$ on an interval, the Mean Value Theorem implies that $y$ is constant on that interval. The initial value forces that constant to be zero. Hence $\boxed{y\equiv 0\text{ is the unique IVP solution.}}$

Step 4: Smoothness versus a series representation. Smoothness means the derivatives of all orders exist and are continuous. A function is analytic at a point if it equals its Taylor series in some neighborhood of that point. The supplied $v$ is smooth but not analytic at $0$: its Taylor series is zero, whereas $v(x)>0$ for every $x>0$. Agreement of derivatives at a single point therefore does not replace the interval identity required by the ODE.