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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 9, Problema 9.3.43a

42–43. Implicit solutions for separable equations For the following separable equations, carry out the indicated analysis.
a. Find the general solution of the equation.


e⁻ʸᐟ²y'(x) = 4x sin x² − x; y(0) = 0, y(0) = ln(1/4), y(√(π/2)) = 0


Guida verificata passo dopo passo
1
Rewrite the given differential equation \(e^{-\frac{y}{2}} y'(x) = 4x \sin(x^2) - x\) in the form \(\frac{dy}{dx} = e^{\frac{y}{2}} (4x \sin(x^2) - x)\) to isolate \(y'\) on one side.
Recognize that the equation is separable, so rearrange terms to separate variables: \(e^{-\frac{y}{2}} dy = (4x \sin(x^2) - x) dx\).
Integrate both sides: \(\int e^{-\frac{y}{2}} dy = \int (4x \sin(x^2) - x) dx\). For the left side, use substitution \(u = -\frac{y}{2}\); for the right side, split the integral into two parts and use substitution for \(\int 4x \sin(x^2) dx\).
After integrating, include the constant of integration \(C\) and write the implicit general solution relating \(y\) and \(x\).
Use the given initial conditions \(y(0) = 0\), \(y(0) = \ln(\frac{1}{4})\), and \(y(\sqrt{\frac{\pi}{2}}) = 0\) to solve for the constant \(C\) and verify the solution's consistency.

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Separable Differential Equations

A separable differential equation can be written as the product of a function of x and a function of y, allowing the variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently, facilitating the solution of the differential equation.
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Solving Separable Differential Equations

Implicit Solutions

An implicit solution to a differential equation is a relation involving both x and y that defines y implicitly as a function of x. Unlike explicit solutions, implicit solutions may not isolate y on one side but still satisfy the differential equation and initial conditions.
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Finding The Implicit Derivative

Initial Conditions and Particular Solutions

Initial conditions specify the value of the solution at a particular point, allowing determination of the constant of integration in the general solution. Applying these conditions yields a unique particular solution that fits the given problem context.
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Initial Value Problems
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a. Show that the logistic growth rate function f(P)=rP(1−P/K) has a maximum value of rK/4 at the point P=K/2.

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a. For the following values of v₀ and s₀, find the position and velocity functions for all times at which the object is above the ground (s = 0).

v₀ = 49 m/s, s₀ = 60 m

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a. Find the approximations to y(0.2) and y(0.4) using Euler’s method with time steps of Δt = 0.2, 0.1, 0.05, and 0.025.


y′(t) = y/2, y(0) = 2; y(t) = 2eᵗᐟ²

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33–36. {Use of Tech} Computing Euler approximations Use a calculator or computer program to carry out the following steps.

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