Domanda del libro di testo
5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.
y'(t) = eʸᐟ²sin t
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Ch. 9 - Differential Equations
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.3.24
17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.
y'(t) = cos² y, y(1) = π/4
Guida verificata passo dopo passo1
First, rewrite the differential equation in Leibniz notation: \(\frac{dy}{dt} = \cos^{2} y\).
Check if the equation is separable by expressing it as a product of a function of \(y\) and a function of \(t\). Here, rewrite as \(\frac{dy}{dt} = \cos^{2} y = f(y) \cdot g(t)\), where \(f(y) = \cos^{2} y\) and \(g(t) = 1\).
Since the equation is separable, separate variables by dividing both sides by \(\cos^{2} y\) and multiplying both sides by \(dt\): \(\frac{1}{\cos^{2} y} dy = dt\).
Integrate both sides: \(\int \frac{1}{\cos^{2} y} dy = \int dt\). Recall that \(\frac{1}{\cos^{2} y} = \sec^{2} y\), and the integral of \(\sec^{2} y\) with respect to \(y\) is \(\tan y\).
After integrating, apply the initial condition \(y(1) = \frac{\pi}{4}\) to solve for the constant of integration and express the solution implicitly or explicitly in terms of \(y\) and \(t\).
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Separable Differential Equations
A differential equation is separable if it can be written as a product of a function of t and a function of y, 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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06:06
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 fixing the constant of integration.
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Percorso guidato
05:03Initial Value Problems
Integration Techniques for Trigonometric Functions
Solving differential equations involving trigonometric functions often requires using identities and integration methods, such as rewriting cos² y using power-reduction formulas. Mastery of these techniques is essential to integrate and solve the equation explicitly.
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Percorso guidato
6:04Introduction to Trigonometric Functions
Pratica correlata
Domanda del libro di testo
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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Domanda del libro di testo
What is a separable first-order differential equation?
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Domanda del libro di testo
33–42. Solving initial value problems Solve the following initial value problems.
y'(t) = 1 + eᵗ, y(0) = 4
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Domanda del libro di testo
9–14. Growth rate functions Make a sketch of the population function P (as a function of time) that results from the following growth rate functions. Assume the population at time t = 0 begins at some positive value.
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Domanda del libro di testo
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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