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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.3.12
Chapter 9, Problem 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)

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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.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.
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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.
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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.
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Divergence Test (nth Term Test)
Related Practice
Textbook Question

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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Textbook Question

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

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Textbook Question

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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Textbook Question

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

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Textbook Question

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.

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Textbook Question

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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