Case 2 of the general solution Solve the equation y′(t) = ky + b in the case that ky + b \u003c 0 and verify that the general solution is y(t) = Ceᵏᵗ − b/k.

Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 9 - Differential Equations

Problem 9.4.36

Chapter 9, Problem 9.4.36

Case 2 of the general solution Solve the equation y′(t) = ky + b in the case that ky + b < 0 and verify that the general solution is y(t) = Ceᵏᵗ − b/k.

Verified step by step guidance

Recognize that the given differential equation is a first-order linear ordinary differential equation of the form \(y'(t) = ky + b\), where \(k\) and \(b\) are constants.

Rewrite the equation as \(y'(t) - ky = b\) to identify it in the standard linear form \(y' + p(t)y = q(t)\), where \(p(t) = -k\) and \(q(t) = b\).

Find the integrating factor \(\mu(t)\), which is given by \(\mu(t) = e^{\int -k \, dt} = e^{-kt}\).

Multiply both sides of the differential equation by the integrating factor to get \(e^{-kt} y'(t) - k e^{-kt} y = b e^{-kt}\), which simplifies to \(\frac{d}{dt} \left( e^{-kt} y \right) = b e^{-kt}\).

Integrate both sides with respect to \(t\): \(\int \frac{d}{dt} \left( e^{-kt} y \right) dt = \int b e^{-kt} dt\), then solve for \(y(t)\) to find the general solution in the form \(y(t) = C e^{kt} - \frac{b}{k}\), where \(C\) is the constant of integration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

First-Order Linear Differential Equations

These are differential equations of the form y' = ky + b, where k and b are constants. They describe rates of change proportional to the function itself plus a constant term. Understanding their structure is essential for finding general solutions.

General Solution of Non-Homogeneous Equations

The general solution combines the homogeneous solution (y' = ky) and a particular solution to the non-homogeneous equation (y' = ky + b). This approach ensures all possible solutions are captured, typically expressed as y(t) = Ce^{kt} plus a constant term.

Verification by Substitution

To verify a proposed solution, substitute it back into the original differential equation. This confirms whether the solution satisfies the equation for all t, ensuring correctness and understanding of the solution's behavior.

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