C: none of these are solutions to the given equation.
• If y(x) = e², then y is constant and y' = 0. Then y' - y = -e² ≠ 0.
• If y(x) = x, then y' = 1, but y' - y = 1 - x ≠ 0.
The actual solution is easy to find, since this equation is separable.
y' - y = 0
dy/dx = y
dy/y = dx
∫ dy/y = ∫ dx
ln|y| = x + C
y = exp(x + C )
y = C exp(x) = C eˣ
