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) = cos² y, y(1) = π/4
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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 9.3.36
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
Guida verificata passo dopo passo1
Rewrite the given differential equation in terms of \( y \) and \( x \). The equation is \( y \frac{dy}{dx} = \frac{2x}{(2 + y)^2} \). Our goal is to separate variables if possible.
Express \( \frac{dy}{dx} \) explicitly: \( \frac{dy}{dx} = \frac{2x}{y(2 + y)^2} \). This suggests separating variables by bringing all \( y \)-terms to one side and \( x \)-terms to the other.
Rewrite the equation as \( y(2 + y)^2 dy = 2x dx \). This sets up the integral form where we integrate both sides with respect to their variables.
Integrate both sides: \( \int y(2 + y)^2 dy = \int 2x dx \). You will need to expand \( (2 + y)^2 \) and then multiply by \( y \) before integrating the left side.
After integrating, apply the initial condition \( y(1) = -1 \) to solve for the constant of integration. The resulting equation will be an implicit solution involving \( x \) and \( y \).
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Implicit Differentiation and Implicit Solutions
Implicit differentiation involves finding derivatives when the function is not explicitly solved for one variable. In this problem, the solution is left in implicit form, meaning y is defined by an equation involving both x and y, rather than y expressed explicitly as a function of x.
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Percorso guidato
05:14
Finding The Implicit Derivative
Initial Value Problems (IVPs)
An initial value problem specifies a differential equation along with a condition that the solution must satisfy at a particular point. This condition, y(1) = -1, helps identify the unique solution curve among potentially many implicit solutions.
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Percorso guidato
05:03Initial Value Problems
Use of Technology for Graphing Implicit Solutions
Graphing software can plot implicit solutions that are difficult to solve explicitly. It helps visualize the solution curve and distinguish which branch corresponds to the initial condition, especially when the implicit equation defines multiple functions.
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Percorso guidato
09:16Finding The Implicit Derivative Example 1
Pratica correlata
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
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) = sec² t/(2y), y(π/4) = 1
45
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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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