Explain why or why not Determine whether the following statements are true and give an explanation or counterexample


b. If k>0 and b>0 then y(t)=0 is never a solution of y'(t)=ky−b.

Verified step by step guidance

1. Start by writing down the differential equation: \(y'(t) = k y - b\), where \(k > 0\) and \(b > 0\).
2. Check if \(y(t) = 0\) can be a solution by substituting \(y(t) = 0\) into the differential equation: compute \(y'(t)\) when \(y(t) = 0\).
3. Substitute \(y(t) = 0\) into the right-hand side: \(k \cdot 0 - b = -b\). Since \(b > 0\), this equals \(-b < 0\).
4. Note that if \(y(t) = 0\) were a solution, then \(y'(t)\) would have to be equal to the right-hand side evaluated at \(y=0\), but \(y'(t)\) would be zero (since \(y(t)\) is constant zero), which contradicts \(y'(t) = -b\).
5. Conclude that \(y(t) = 0\) is not a solution of the differential equation when \(k > 0\) and \(b > 0\), because the derivative condition is not satisfied.

Key Concepts

Differential Equations and Their Solutions

A differential equation relates a function and its derivatives. Solutions are functions that satisfy the equation when substituted. Understanding how to verify if a function is a solution involves substituting it into the equation and checking if both sides are equal.

Equilibrium (Constant) Solutions

Equilibrium solutions occur when the derivative equals zero, meaning the function is constant over time. For y'(t) = ky - b, setting y'(t) = 0 helps find constant solutions y = b/k. Recognizing these helps determine if y(t) = 0 can be a solution.

Substitution and Counterexamples

To test if y(t) = 0 is a solution, substitute y = 0 into the differential equation. If the resulting equality holds, it is a solution; otherwise, it is not. Providing a counterexample involves showing a specific case where the statement fails.