Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. The general solution of the differential equation y'(t) = 1 is y(t) = t

Verified step by step guidance

1. Identify the given differential equation: \(y'(t) = 1\). This means the derivative of \(y\) with respect to \(t\) is constantly 1.
2. Recall that the general solution to a differential equation of the form \(y'(t) = c\), where \(c\) is a constant, is \(y(t) = ct + C\), where \(C\) is an arbitrary constant of integration.
3. Apply this to the given equation: since \(c = 1\), the general solution should be \(y(t) = 1 \cdot t + C\), or simply \(y(t) = t + C\).
4. Compare this with the proposed solution \(y(t) = t\). Notice that the proposed solution does not include the constant of integration \(C\).
5. Conclude that the statement is false because the general solution must include the constant \(C\) to account for all possible solutions, not just \(y(t) = t\).

Key Concepts

General Solution of a Differential Equation

The general solution of a differential equation includes all possible solutions and typically contains an arbitrary constant. It represents a family of functions that satisfy the equation, not just a single function.

Integration of Derivatives

To find the general solution of y'(t) = 1, integrate the right-hand side with respect to t. Since the derivative of y is 1, integrating gives y(t) = t + C, where C is an arbitrary constant.

Counterexample and Verification

To verify if a proposed solution is general, check if it includes all solutions. The statement y(t) = t omits the constant C, so it is not the general solution. A counterexample is y(t) = t + 5, which also satisfies y'(t) = 1.