Textbook Question
5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.
y'(t) = eʸᐟ²sin t
Ch. 9 - Differential Equations
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- 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: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 9.2.2
Consider the differential equation y'(t) = t² - 3y² and the solution curve that passes through the point (3, 1). What is the slope of the curve at (3, 1)?
Verified step by step guidance1
Identify the given differential equation: \(y'(t) = t^{2} - 3y^{2}\).
Recognize that \(y'(t)\) represents the slope of the solution curve at any point \((t, y)\).
Substitute the given point \((t, y) = (3, 1)\) into the differential equation to find the slope at that point.
Calculate the value of the right-hand side by plugging in \(t = 3\) and \(y = 1\): \(3^{2} - 3 \times 1^{2}\).
Simplify the expression to find the slope \(y'(3)\) at the point \((3, 1)\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Differential Equation
A differential equation relates a function with its derivatives, describing how the function changes. In this problem, y'(t) represents the derivative of y with respect to t, indicating the slope of the solution curve at any point.
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Evaluating the Derivative at a Point
To find the slope of the solution curve at a specific point, substitute the given values of t and y into the differential equation. This gives the instantaneous rate of change or slope at that point.
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Initial Condition and Solution Curve
The initial condition (3, 1) specifies a point through which the solution curve passes. It helps identify the particular solution of the differential equation and allows evaluation of the slope at that exact point.
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Related Practice
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.
yy'(x) = 2x/(2 + y)², y(1) = −1
Textbook Question
20–22. {Use of Tech} Solving the Gompertz equation Solve the Gompertz equation in Exercise 19 with the given values of r, K, and M₀. Then graph the solution to be sure that M(0) and lim(t→∞) M(t) are correct.
r = 0.05, K = 1200, M₀ = 90
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Textbook Question
What is the equilibrium solution of the equation y'(t) = 3y − 9? Is it stable or unstable?
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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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Textbook Question
17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.
y(t) = sec² t/(2y), y(π/4) = 1
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