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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.1.12

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

Guida verificata passo dopo passo
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.
Compute the first derivative \(u'(t)\) using the product rule for the term \(C_2 t e^t\): \(u'(t) = \frac{d}{dt}(C_1 e^t) + \frac{d}{dt}(C_2 t e^t) = C_1 e^t + C_2 \left( e^t + t e^t \right)\).
Simplify the first derivative: \(u'(t) = C_1 e^t + C_2 e^t + C_2 t e^t = (C_1 + C_2) e^t + C_2 t e^t\).
Compute the second derivative \(u''(t)\) by differentiating \(u'(t)\) again, applying the product rule to the \(C_2 t e^t\) term: \(u''(t) = \frac{d}{dt} \left( (C_1 + C_2) e^t + C_2 t e^t \right) = (C_1 + C_2) e^t + C_2 \left( e^t + t e^t \right)\).
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.

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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.
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Solutions to Basic Differential Equations

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.
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Substitution With an Extra Variable

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.
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Derivatives of General Exponential Functions
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