Basic Concepts for nth Order Linear Equations — Question 8

Question 8 – Solution

Strategy. Use the full initial derivative vector, rather than only the value and slope, to apply uniqueness.

Step 1: Verify a fundamental set. Each candidate satisfies $y'''+y'=0$. Their Wronskian is $$W=\det\begin{pmatrix}1&\sin x&\cos x\\0&\cos x&-\sin x\\0&-\sin x&-\cos x\end{pmatrix} =\boxed{-1}.$$ Thus the three solutions are independent. The regular third-order equation has a three-dimensional homogeneous solution space, so they form a fundamental set.

Step 2: Determine the unique IVP solution. Write $y=A+B\sin x+C\cos x$. The data give $A+C=0$, $B=0$ and $-C=1$. Hence $$\boxed{y=1-\cos x.}$$ Its first two derivatives are $\sin x$ and $\cos x$, giving initial vector $(0,0,1)$; its third derivative $-\sin x$ verifies the equation.

Step 3: Classify all zero contacts. The equation $1-\cos x=0$ holds exactly at $x=2k\pi$, $k\in\mathbb Z$. At each such point, $y'=0$ but $y''=1\ne 0$, so every zero has multiplicity exactly two. There are infinitely many, with no finite accumulation point. Their repeated tangency does not make the function identically zero.

Step 4: State the correct uniqueness consequence. For a normalized homogeneous linear equation of order $n$ with continuous coefficients on a connected interval, if $$\boxed{y(x_0)=y'(x_0)=\cdots=y^{(n-1)}(x_0)=0,}$$ then uniqueness of the zero-data IVP forces $y\equiv 0$ throughout that interval. Here $n=3$, and the second derivative at every zero is $1$, so the required three-component vector is not zero. Value and slope alone are insufficient; an arbitrary number of separated double zeros is not excluded by the IVP theorem.