Questions: Separable Equations (Intro)

Question 1

After solving dy/dx = 3y by separation of variables, you get y = Ce^(3x). What does the constant C represent?

• A. The rate of growth of y with respect to x
• B. An arbitrary constant determined by initial conditions
• C. The value of y when x = 0, always equal to 1
• D. A constant that cancels when you differentiate to verify the solution

Answer: (B)

C arises from the constant of integration and is determined by an initial condition. If you know y(0) = 5, then 5 = Ce^0 = C, so C = 5. C does not cancel when differentiating — d/dx(Ce^(3x)) = 3Ce^(3x) = 3y, which confirms the solution. The constant encodes the starting value of the system.

Question 2

When solving a separable equation dy/dx = f(x)g(y), you can usually safely divide both sides by g(y) before integrating.

• T. True
• F. False

Answer: False

Division by g(y) is only valid when g(y) ≠ 0. If g(y) = 0 for some constant value y = c, then y = c is an equilibrium solution that the separation-of-variables process will miss entirely. Always check whether g(y) = 0 has solutions and handle them separately.

Question 3

Why does integrating 1/y with respect to y produce ln|y| rather than ln(y), and when is it valid to drop the absolute value?

Model answer: The integral of 1/y is ln|y| because the natural log is only defined for positive arguments, and y could be negative depending on the problem. The absolute value can be dropped when the context guarantees y > 0 — for example, when y represents a population size or a physical quantity that cannot be negative — or when applying an initial condition confirms the sign.

Skipping the absolute value introduces subtle errors: for instance, writing ln(y) = x + C and then exponentiating gives y = e^(x+C) = Ae^x, which only captures positive solutions. The correct form ln|y| = x + C gives y = ±e^(x+C) = Ae^x where A can be any nonzero constant, including negative. The absolute value keeps track of sign information.