Domanda del libro di testo
Explain how the growth rate function determines the solution of a population model.
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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 9.3.30
17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.
y'(t) = y³sin t, y(0) = 1
Guida verificata passo dopo passo1
Identify whether the differential equation is separable. The given equation is \(y'(t) = y^{3} \sin t\). Since the right side can be expressed as a product of a function of \(y\) and a function of \(t\), it is separable.
Rewrite the differential equation using Leibniz notation: \(\frac{dy}{dt} = y^{3} \sin t\). Then separate variables by dividing both sides by \(y^{3}\) and multiplying both sides by \(dt\): \(\frac{1}{y^{3}} dy = \sin t \, dt\).
Integrate both sides: \(\int \frac{1}{y^{3}} dy = \int \sin t \, dt\). This will give you two antiderivatives, one in terms of \(y\) and one in terms of \(t\), plus a constant of integration.
Solve the integrals: Recall that \(\int y^{-3} dy = \int y^{-3} dy\) and \(\int \sin t \, dt = -\cos t + C\). Write the integrated form explicitly, including the constant of integration on one side.
Apply the initial condition \(y(0) = 1\) to find the constant of integration. Substitute \(t=0\) and \(y=1\) into the integrated equation and solve for the constant. Then express \(y\) explicitly as a function of \(t\) if possible.
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Separable Differential Equations
A differential equation is separable if it can be written 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, simplifying the solution process.
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06:06
Solving Separable Differential Equations
Initial Value Problems (IVP)
An initial value problem specifies the value of the unknown function at a particular point, providing a unique solution to a differential equation. Solving an IVP involves finding the general solution and then applying the initial condition to determine the constant of integration.
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Percorso guidato
05:03Initial Value Problems
Integration Techniques for Solving ODEs
Once variables are separated, integration is used to solve the resulting expressions. This often involves integrating standard functions like powers of y and trigonometric functions of t. Proper integration and algebraic manipulation yield the explicit or implicit solution to the differential equation.
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Percorso guidato
06:18Integration by Parts for Definite Integrals
Pratica correlata
Domanda del libro di testo
7–16. Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume C, C1, C2 and C3 are arbitrary constants.
u(t) = C₁eᵗ + C₂teᵗ; u''(t) - 2u'(t) + u(t) = 0
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Domanda del libro di testo
7–16. Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume C, C1, C2 and C3 are arbitrary constants.
y(t) = C₁ sin4t + C₂ cos4t; y''(t) + 16y(t) = 0
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Domanda del libro di testo
5–10. First-order linear equations Find the general solution of the following equations.
v'(y) − v/2 = 14
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Domanda del libro di testo
45–48. General first-order linear equations Consider the general first-order linear equation y'(t)+a(t)y(t)=f(t). This equation can be solved, in principle, by defining the integrating factor p(t)=exp(∫a(t)dt). Here is how the integrating factor works. Multiply both sides of the equation by p (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes
p(t)(y′(t) + a(t)y(t)) = d/dt(p(t)y(t)) = p(t)f(t).
Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor.
y′(t) + (2t)/(t² + 1)y(t) = 1 + 3t², y(1) = 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) = t lnt + 1
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