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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
Tutti i libri di testoBriggs 3rd EditionCh. 9 - Differential EquationsProblema 9.3.27
Capitolo 9, Problema 9.3.27

17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.
y(t) = sec² t/(2y), y(π/4) = 1

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First, rewrite the given differential equation in the form \( \frac{dy}{dt} = \frac{\sec^{2} t}{2y} \). This makes it clear that \( y \) is a function of \( t \) and \( \frac{dy}{dt} \) is expressed in terms of \( t \) and \( y \).
Check if the equation is separable by trying to express it as \( g(y) dy = f(t) dt \). Multiply both sides by \( 2y \) to get \( 2y \frac{dy}{dt} = \sec^{2} t \), then rewrite as \( 2y dy = \sec^{2} t dt \). This shows the variables can be separated.
Integrate both sides: integrate \( 2y dy \) with respect to \( y \) and integrate \( \sec^{2} t dt \) with respect to \( t \). This will give you two antiderivatives, one in terms of \( y \) and one in terms of \( t \), plus a constant of integration.
After integration, you will have an implicit solution involving \( y \) and \( t \). Use the initial condition \( y(\pi/4) = 1 \) to solve for the constant of integration.
Finally, express \( y \) explicitly as a function of \( t \) if possible, or leave the solution in implicit form if it cannot be easily solved for \( y \).

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Separable Differential Equations

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

Initial Value Problems (IVP)

An initial value problem involves solving a differential equation with a given initial condition, such as y(t₀) = y₀. This condition helps determine the specific solution curve among the family of solutions by finding the constant of integration.
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Initial Value Problems

Integration Techniques for Solving ODEs

Once variables are separated, integration is used to solve the resulting equations. This may involve standard integration methods or recognizing common integrals, such as those involving trigonometric functions, to find an explicit or implicit solution.
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Integration by Parts for Definite Integrals
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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.

yy'(x) = 2x/(2 + y)², y(1) = −1

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What is a separable first-order differential equation?

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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.

y'(t) = 3 + e⁻²ᵗ

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Domanda del libro di testo

17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.

y'(t) = eᵗʸ, y(0) = 1

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Domanda del libro di testo

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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12–16. Sketching direction fields Use the window [-2, 2] x [-2, 2] to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. A detailed direction field is not needed.

y'(t) = 4−y, y(0) = −1

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