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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21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.

p'(x) = 16/x⁹ - 5 + 14x⁶

144

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33–38. {Use of Tech} Solutions in implicit form Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one function, be sure to indicate which function corresponds to the solution of the initial value problem.

z(x) = (z² + 4)/(x² + 16), z(4) = 2

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Explain why the graph of the solution to the initial value problem y'(t) = t²/(1 - t), y(-1) = ln 2 cannot cross the line t = 1.

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5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.

(t² + 1)³yy'(t) = t(y² + 4)

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y'(t) = 2t²/(y² − 1), y(0) = 0

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Textbook Question

5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.

e⁴ᵗy'(t) = 5

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