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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
Chapter 9, Problem 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

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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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Key Concepts

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

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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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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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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Related Practice
Textbook Question

Explain how the growth rate function determines the solution of a population model.

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Textbook Question

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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Textbook Question

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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Textbook Question

5–10. First-order linear equations Find the general solution of the following equations.


v'(y) − v/2 = 14

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Textbook Question

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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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) = t lnt + 1

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