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.
p'(x) = 16/x⁹ - 5 + 14x⁶
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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 9.3.38
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.
z(x) = (z² + 4)/(x² + 16), z(4) = 2
Guida verificata passo dopo passo1
Rewrite the given differential equation in the form \( \frac{dz}{dx} = \frac{z^2 + 4}{x^2 + 16} \) to clearly identify the derivative \( \frac{dz}{dx} \).
Separate the variables by rearranging terms to isolate \( z \) on one side and \( x \) on the other: \( \frac{dz}{z^2 + 4} = \frac{dx}{x^2 + 16} \).
Integrate both sides: compute \( \int \frac{dz}{z^2 + 4} \) and \( \int \frac{dx}{x^2 + 16} \). Recall that \( \int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C \).
After integrating, write the implicit solution by equating the two antiderivatives plus a constant of integration \( C \): \( \frac{1}{2} \arctan\left(\frac{z}{2}\right) = \frac{1}{4} \arctan\left(\frac{x}{4}\right) + C \).
Use the initial condition \( z(4) = 2 \) to solve for the constant \( C \), then express the implicit solution including this constant. This implicit equation represents the solution to the initial value problem.
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Implicit Solutions of Differential Equations
An implicit solution is a relation involving both the dependent and independent variables that satisfies the differential equation without explicitly solving for the dependent variable. Such solutions may define multiple functions, and additional conditions help identify the specific solution relevant to the initial value problem.
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05:09
Verifying Solutions of Differential Equations
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 allows us to determine the unique solution curve among many possible implicit or explicit solutions.
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Percorso guidato
05:03Initial Value Problems
Use of Technology for Graphing Solutions
Graphing software helps visualize implicit solutions that are difficult to solve explicitly. By plotting the implicit relation and applying the initial condition, one can identify the correct solution branch and better understand the behavior of the solution.
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Percorso guidato
06:15Graphing The Derivative
Pratica correlata
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.
(t² + 1)³yy'(t) = t(y² + 4)
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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.
y'(t) = 2t²/(y² − 1), y(0) = 0
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Domanda del libro di testo
21–24. Logistic equations Consider the following logistic equations. In each case, sketch the direction field, draw the solution curve for each initial condition, and find the equilibrium solutions. A detailed direction field is not needed. Assume t ≥ 0 and tP ≥ 0.
P′(t) = 0.05P − 0.001P²; P(0) = 10, P(0) = 40, P(0) = 80
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Domanda del libro di testo
Case 2 of the general solution Solve the equation y′(t) = ky + b in the case that ky + b < 0 and verify that the general solution is y(t) = Ceᵏᵗ − b/k.
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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.
e⁴ᵗy'(t) = 5
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