7–16. Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume C, C1, C2 and C3 are arbitrary constants.
u(t) = C₁eᵗ + C₂teᵗ; u''(t) - 2u'(t) + u(t) = 0

Verified step by step guidance

1. Identify the given function: \(u(t) = C_1 e^t + C_2 t e^t\), where \(C_1\) and \(C_2\) are arbitrary constants.
2. Compute the first derivative \(u'(t)\) using the product rule for the term \(C_2 t e^t\): \(u'(t) = \frac{d}{dt}(C_1 e^t) + \frac{d}{dt}(C_2 t e^t) = C_1 e^t + C_2 \left( e^t + t e^t \right)\).
3. Simplify the first derivative: \(u'(t) = C_1 e^t + C_2 e^t + C_2 t e^t = (C_1 + C_2) e^t + C_2 t e^t\).
4. Compute the second derivative \(u''(t)\) by differentiating \(u'(t)\) again, applying the product rule to the \(C_2 t e^t\) term: \(u''(t) = \frac{d}{dt} \left( (C_1 + C_2) e^t + C_2 t e^t \right) = (C_1 + C_2) e^t + C_2 \left( e^t + t e^t \right)\).
5. Substitute \(u(t)\), \(u'(t)\), and \(u''(t)\) into the differential equation \(u''(t) - 2 u'(t) + u(t) = 0\) and simplify the expression. If the left-hand side simplifies to zero for all \(t\), then \(u(t)\) is a solution to the differential equation.

Key Concepts

General Solution of a Differential Equation

The general solution of a differential equation includes all possible solutions and typically contains arbitrary constants. It represents the complete set of functions that satisfy the equation, allowing for initial conditions to specify a unique solution.

Verification by Substitution

To verify a solution, substitute the given function and its derivatives into the differential equation. If the equation holds true for all values in the domain, the function is a valid solution.

Derivatives of Exponential Functions

Understanding how to compute derivatives of functions involving exponentials and products, such as te^t, is essential. Use the product rule for derivatives when differentiating terms like C₂teᵗ.