Skip to main content
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 9.3.12
Capitolo 9, Problema 9.3.12

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)

Guida verificata passo dopo passo
1
Rewrite the given differential equation \((t^2 + 1)^3 y y'(t) = t (y^2 + 4)\) by expressing \(y'(t)\) as \(\frac{dy}{dt}\), so it becomes \((t^2 + 1)^3 y \frac{dy}{dt} = t (y^2 + 4)\).
Separate the variables by moving all terms involving \(y\) to one side and all terms involving \(t\) to the other side. This gives \(\frac{y}{y^2 + 4} dy = \frac{t}{(t^2 + 1)^3} dt\).
Integrate both sides: compute \(\int \frac{y}{y^2 + 4} dy\) on the left and \(\int \frac{t}{(t^2 + 1)^3} dt\) on the right.
For the left integral, use substitution \(u = y^2 + 4\), so $du = 2y dy$, which simplifies the integral. For the right integral, consider substitution \(v = t^2 + 1\), so $dv = 2t dt$, to simplify the integral.
After integrating both sides, include the constant of integration \(C\), then solve the resulting equation explicitly for \(y\) as a function of \(t\) to find the general solution.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
6m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Separable Differential Equations

A separable differential equation can be written as a product of a function of the dependent variable and a function of the independent variable. This allows the equation to be rearranged so that all terms involving one variable are on one side and all terms involving the other variable are on the opposite side, enabling integration on both sides.
Video consigliato:
06:06
Solving Separable Differential Equations

Implicit and Explicit Solutions

An implicit solution defines the dependent variable in terms of the independent variable without isolating it explicitly, while an explicit solution expresses the dependent variable directly as a function of the independent variable. The problem asks for the explicit form, so after integration, solving for the dependent variable is necessary.
Video consigliato:
Percorso guidato
05:14
Finding The Implicit Derivative

Integration Techniques for Nonlinear Terms

Solving separable equations often requires integrating nonlinear expressions involving powers or polynomials. Familiarity with integration rules, substitution, and algebraic manipulation is essential to handle terms like (t² + 1)³ and y² + 4, ensuring correct integration and simplification.
Video consigliato:
Percorso guidato
05:44
Divergence Test (nth Term Test)
Pratica correlata
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

144
views
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.

z(x) = (z² + 4)/(x² + 16), z(4) = 2

43
views
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

60
views
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

51
views
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.

53
views
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.

63
views