Definitions — Question 10

Question 10 – Solution

Strategy. Determine when the leading coefficient or the nonlinear coefficient vanishes. Classify only after substituting those exceptional parameter values.

Step 1: Exhaustive classification. Factor $a^2-1=(a-1)(a+1)$. $$\begin{array}{c|c|l} a&\text{Simplified equation}&\text{Classification}\\\hline 1&2y'=x&\text{First order, linear}\\ -1&-2y''=x&\text{Second order, linear}\\ a\ne\pm 1&\text{Original equation}&\text{Second order, nonlinear} \end{array}$$ For $a\ne\pm 1$, both the $y''$ coefficient and the $(y')^2$ coefficient are nonzero. Both linear exceptional equations are nonhomogeneous because their right-hand sides are not identically zero.

Step 2: Solve when $a=1$. From $y'=x/2$, $$y=\frac{x^2}{4}+C,\qquad y(0)=0\Longrightarrow C=0.$$ Thus $\boxed{y=x^2/4\text{ on }\mathbb R}$. Its derivative at $0$ is $0$, as required, and $2y'=x$ verifies the equation directly.

Step 3: Solve when $a=-1$. Two integrations of $y''=-x/2$ give $$y'=-\frac{x^2}{4}+C_1,\qquad y=-\frac{x^3}{12}+C_1x+C_2.$$ The data yield $C_1=C_2=0$, so $\boxed{y=-x^3/12\text{ on }\mathbb R}$. Indeed $-2y''=x$ and both initial values are $0$.

Step 4: Redundant or incompatible data. For $a=1$, the equation at $x=0$ already forces $2y'(0)=0$. Thus the stated slope condition is redundant; replacing it with $y'(0)=1$ makes the problem inconsistent. The coefficient $a-1$ vanishes at $a=1$, removing $y''$ entirely. Order is determined by derivatives with nonzero coefficients in the actual equation, not by unsimplified notation.