Definitions — Question 6

Question 6 – Solution

Strategy. Use the derivative to remove the constant, then investigate what information was lost and whether new solutions are admitted.

Step 1: Eliminate the parameter. Differentiating yields $y'=2(x-C)$. Squaring and using $y=(x-C)^2$ gives $$\boxed{(y')^2=4y.}$$ This is a first-order nonlinear ODE. The exponent on $y'$ changes linearity, not derivative order.

Step 2: Direct verification. For any real $C$, $$(y')^2=[2(x-C)]^2=4(x-C)^2=4y.$$ The family members are polynomials and therefore satisfy the equation on all of $\mathbb R$, including $x=C$.

Step 3: Apply the initial value. At $x=0$ we require $C^2=1$, so $C=\pm 1$. The two selected family members are $$\boxed{y=(x-1)^2\quad\text{and}\quad y=(x+1)^2.}$$ Their slopes at $0$ are $-2$ and $2$, respectively. The equation itself permits either sign because it specifies $(y')^2$, not $y'$.

Step 4: A solution outside the family. The function $y\equiv 0$ satisfies $(y')^2=4y$ identically. If $(x-C)^2$ were identically $0$ on a nonempty open interval, every $x$ in that interval would have to equal one fixed number $C$, which is impossible. Thus $$\boxed{y\equiv 0\text{ is a solution not contained in the displayed family.}}$$ Elimination proves that family members solve the equation; it does not prove the converse. In particular, a parameterized family cannot be called a complete description merely because substitution succeeds.