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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 9.3.42a

42–43. Implicit solutions for separable equations For the following separable equations, carry out the indicated analysis.
a. Find the general solution of the equation.


y'(t) = t²/(y² + 1); y(−1) = 1, y(0) = 0, y(−1) = −1


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1
Identify that the given differential equation is separable: \(y'(t) = \frac{t^2}{y^2 + 1}\).
Rewrite the equation in separable form by expressing \(y'\) as \(\frac{dy}{dt}\) and rearranging terms to isolate \(y\) and \(t\): $ (y^2 + 1) dy = t^2 dt$.
Integrate both sides separately: \(\int (y^2 + 1) dy = \int t^2 dt\).
Compute the integrals: The left side becomes \(\int y^2 dy + \int 1 dy = \frac{y^3}{3} + y\), and the right side becomes \(\frac{t^3}{3} + C\), where \(C\) is the constant of integration.
Write the implicit general solution as \(\frac{y^3}{3} + y = \frac{t^3}{3} + C\). Use the initial conditions \(y(-1) = 1\), \(y(0) = 0\), and \(y(-1) = -1\) to solve for the constant \(C\) for each case.

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Separable Differential Equations

A separable differential equation can be written as the product of a function of t and a function of y, allowing the variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently to find the general solution.
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Solving Separable Differential Equations

Implicit Solutions

An implicit solution is a relation involving both variables that satisfies the differential equation but is not explicitly solved for y. Often, after integrating, the solution is given implicitly, requiring further manipulation or interpretation to understand the behavior of y.
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Finding The Implicit Derivative

Initial Conditions and Solution Curves

Initial conditions specify particular values of y at given t, allowing determination of the constant of integration and thus a unique solution curve. The direction field or slope field graphically represents these solutions, showing how different initial values lead to different solution trajectories.
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Initial Value Problems
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38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.

a. Find the equilibrium solutions. 


y′(t) = y(y - 3)

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a. Assume t = 0 corresponds to 2005 and that the population growth is exponential for the first ten years; that is, between 2005 and 2015, the population is given by P(t) = P(0)exp(rt). Estimate the growth rate r using this assumption.

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A second-order equation Consider the differential equation y''(t) - k²y(t) = 0 where k > 0 is a real number.


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a. Show that t₁/₂ = −1/k ln((T₀ − 2A)/(2(T₀ − A))).

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23–26. Stirred tank reactions For each of the following stirred tank reactions, carry out the following analysis.

a. Write an initial value problem for the mass of the substance.


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