/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 From \(y^{\prime}+y=y^{-1 / 2}\)... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 22

From \(y^{\prime}+y=y^{-1 / 2}\) and \(w=y^{3 / 2}\) we obtain \(\frac{d w}{d x}+\frac{3}{2} w=\frac{3}{2} .\) An integrating factor is \(e^{3 x / 2}\) so that \(e^{3 x / 2} w=\) \(e^{3 x / 2}+c\) or \(y^{3 / 2}=1+c e^{-3 x / 2} .\) If \(y(0)=4\) then \(c=7\) and \(y^{3 / 2}=1+7 e^{-3 x / 2}\).

Short Answer

Expert verified
The solution is \( y^{3/2} = 1 + 7e^{-3x/2} \).

Step by step solution

01

Rewrite the given transformation

We start with the transformation where a substitution is provided. Given the original differential equation: \[ y' + y = y^{-1/2} \]And the substitution \( w = y^{3/2} \), this implies that \( y = (w)^{2/3} \) and thus its derivative with respect to \( x \) becomes \( \, dy/dx = \frac{2}{3} w^{1/3} \cdot dw/dx \). This allows us to express the given substitution in terms of \( w \) and its derivative.
02

Substitute in the Differential Equation

With \( y = (w)^{2/3} \), differentiate both sides to convert the equation:\[ y' = \frac{2}{3} (w)^{-1/3} \cdot \frac{dw}{dx} \] By substituting \( y \) and \( y' \) into the original equation:\[\frac{2}{3} (w)^{-1/3} \frac{dw}{dx} + (w)^{2/3} = ((w)^{2/3})^{-1/2} \] Further simplifying yields:\[\frac{2}{3} \frac{dw}{dx} = (w)^{-1/2} \] Now multiply through by \( (w)^{1/3} \) to give:\[ \frac{d w}{dx} + \frac{3}{2} w = \frac{3}{2} \]
03

Solve using Integrating Factor

The resulting linear first order equation is:\[ \frac{dw}{dx} + \frac{3}{2}w = \frac{3}{2} \]We find the integrating factor, \( \mu(x) = e^{\int \frac{3}{2} \, dx} = e^{3x/2} \).Multiply through by the integrating factor:\[ e^{3x/2}\frac{dw}{dx} + \frac{3}{2}e^{3x/2}w = e^{3x/2} \cdot \frac{3}{2} \]This results in:\[ \frac{d}{dx} (e^{3x/2}w) = e^{3x/2}\cdot\frac{3}{2} \] Integrating both sides:\[ e^{3x/2}w = e^{3x/2} + C \]
04

Solve for Original Function

Simplify the equation obtained by integration:\[ w = 1 + Ce^{-3x/2} \] Using the transformation \( w = y^{3/2} \), we have:\[ y^{3/2} = 1 + Ce^{-3x/2} \] Substitute the initial condition given, \( y(0) = 4 \). Evaluate the expression:\[ 4^{3/2} = 1 + C \] so \( C = 7 \). This leads to the equation:\[ y^{3/2} = 1 + 7e^{-3x/2} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Factor
The integrating factor is a powerful tool for solving linear first-order differential equations. It is especially useful when dealing with non-homogeneous terms. In this particular problem, the equation is in the form: \[ \frac{dw}{dx} + \frac{3}{2}w = \frac{3}{2} \] This is a typical linear differential equation with constant coefficients.
  • An integrating factor, \( \mu(x) \), is used to simplify such equations.
  • It is usually found with the expression: \( e^{\int P(x) \, dx} \), where \( P(x) \) comes from the equation \( \frac{dw}{dx} + P(x)w = g(x) \).
  • For this exercise, \( P(x) = \frac{3}{2} \), leading to the integrating factor \( e^{3x/2} \).
  • This integrating factor transforms the equation into an exact derivative, enabling straightforward integration.
Multiply through by the integrating factor to form a simple expression: \[ \frac{d}{dx} (e^{3x/2}w) = e^{3x/2} \cdot \frac{3}{2} \] This step is crucial because it turns the left-hand side into an easily integrable form.
Initial Value Problems
Initial value problems (IVP) represent scenarios where the value of the solution is specified at a particular point. For this problem, the initial condition given is \( y(0) = 4 \).
  • It provides a specific solution that satisfies both the differential equation and this initial state.
  • After finding a general solution taking the form \( y^{3/2} = 1 + Ce^{-3x/2} \), substituting \( y(0) = 4 \) helps determine \( C \).
  • Calculate \( 4^{3/2} \): The number is 8, which needs to match \( 1 + C \).
  • This gives \( C = 7 \). So, the equation now represents the particular solution that fits the initial condition.
This type of problem is common in scenarios modeling real-world processes where the system's state is known at a starting condition. This allows us to predict the system's future behavior.
Linear Differential Equations
Linear differential equations are a cornerstone of mathematical modeling. They appear in various fields including physics, engineering, and economics. A linear differential equation of the first order typically has the form: \[ \frac{dy}{dx} + P(x)y = g(x) \] Key characteristics include:
  • The dependent variable and its derivatives appear linearly (no powers, products, or functions of \( y \)).
  • These equations can be homogeneous if \( g(x) = 0 \) or non-homogeneous if \( g(x) eq 0 \).
  • The exercise provided becomes linear through substitution: transforming \( y^{\prime} + y = y^{-1/2} \) to \( \frac{dw}{dx} + \frac{3}{2}w = \frac{3}{2} \).
  • Solving involves finding a particular solution that satisfies the initial conditions.
Understanding how to solve linear differential equations is essential for navigating more complex models. This knowledge is foundational to predicting the behavior of natural systems and engineered processes.

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Most popular questions from this chapter

(a) \((i)\) If \(s(t)\) is distance measured down the plane from the highest point, then \(d s / d t=v .\) Integrating \(d s / d t=16 t\) gives \(s(t)=8 t^{2}+c_{2} .\) Using \(s(0)=0\) then gives \(c_{2}=0 .\) Now the length \(L\) of the plane is \(L=50 / \sin 30^{\circ}=100 \mathrm{ft} .\) The time it takes the box to slide completely down the plane is the solution of \(s(t)=100\) or \(t^{2}=25 / 2,\) so \(t \approx 3.54 \mathrm{s}\). \((i i)\) Integrating \(d s / d t=4 t\) gives \(s(t)=2 t^{2}+c_{2} .\) Using \(s(0)=0\) gives \(c_{2}=0,\) so \(s(t)=2 t^{2}\) and the solution of \(s(t)=100\) is now \(t \approx 7.07 \mathrm{s}\). \((\text {iii})\) Integrating \(d s / d t=48-48 e^{-t / 12}\) and using \(s(0)=0\) to determine the constant of integration, we \(\operatorname{obtain} s(t)=48 t+576 e^{-t / 12}-576 .\) With the aid of a CAS we find that the solution of \(s(t)=100,\) or $$\begin{aligned} &100=48 t+576 e^{-t / 12}-576 &\text { or } \quad 0=48 t+576 e^{-t / 12}-676 \end{aligned}$$ is now \(t \approx 7.84 \mathrm{s}\). (b) The differential equation \(m d v / d t=m g \sin \theta-\mu m g \cos \theta\) can be written $$m \frac{d v}{d t}=m g \cos \theta(\tan \theta-\mu)$$. If \(\tan \theta=\mu, d v / d t=0\) and \(v(0)=0\) implies that \(v(t)=0 .\) If \(\tan \theta< \mu\) and \(v(0)=0,\) then integration implies \(v(t)=g \cos \theta(\tan \theta-\mu) t < 0\) for all time \(t\). (c) since \(\tan 23^{\circ}=0.4245\) and \(\mu=\sqrt{3} / 4=0.4330,\) we see that \(\tan 23^{\circ}<0.4330 .\) The differential equation is \(d v / d t=32 \cos 23^{\circ}\left(\tan 23^{\circ}-\sqrt{3} / 4\right)=-0.251493 .\) Integration and the use of the initial condition gives \(v(t)=-0.251493 t+1 .\) When the box stops, \(v(t)=0\) or \(0=-0.251493 t+1\) or \(t=3.976254 \mathrm{s}\). From \(s(t)=-0.125747 t^{2}+t\) we find \(s(3.976254)=1.988119 \mathrm{ft}\). (d) With \(v_{0} >0, v(t)=-0.251493 t+v_{0}\) and \(s(t)=-0.125747 t^{2}+v_{0} t .\) Because two real positive solutions of the equation \(s(t)=100,\) or \(0=-0.125747 t^{2}+v_{0} t-100,\) would be physically meaningless, we use the quadratic formula and require that \(b^{2}-4 a c=0\) or \(v_{0}^{2}-50.2987=0 .\) From this last equality we find \(v_{0} \approx 7.092164 \mathrm{ft} / \mathrm{s} .\) For the time it takes the box to traverse the entire inclined plane, we must have \(0=-0.125747 t^{2}+7.092164 t-100 .\) Mathematica gives complex roots for the last equation: \(t=\) \(28.2001 \pm 0.0124458 i .\) But, for $$0=-0.125747 t^{2}+7.092164691 t-100$$ the roots are \(t=28.1999 \mathrm{s}\) and \(t=28.2004 \mathrm{s} .\) So if \(v_{0} >7.092164,\) we are guaranteed that the box will slide completely down the plane.

For \(\frac{d i}{d t}+\frac{R}{L} i=\frac{E}{L}\) an integrating factor is \(e^{\int(R / L) d t}=e^{R t / L}\) so that \(\frac{d}{d t}\left[e^{R t / L} i\right]=\frac{E}{L} e^{R t / L}\) and \(i=\frac{E}{R}+c e^{-R t / L}\) for \(-\infty< t<\infty .\) If \(i(0)=i_{0}\) then \(c=i_{0}-E / R\) and \(i=\frac{E}{R}+\left(i_{0}-\frac{E}{R}\right) e^{-R t / L}\).

(a) Separating variables and integrating, we have \\[\left(-2 y+y^{2}\right) d y=\left(x-x^{2}\right) d x\\] and \\[-y^{2}+\frac{1}{3} y^{3}=\frac{1}{2} x^{2}-\frac{1}{3} x^{3}+c\\]. Using a CAS we show some contours of \\[f(x, y)=2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}\\]. The plots shown on \([-7,7] \times[-5,5]\) correspond to \(c\) -values of -450,-300,-200,-120,-60,-20,-10,-8.1,-5 \(-0.8,20,60,\) and 120. (b) The value of \(c\) corresponding to \(y(0)=\frac{3}{2}\) is \(f\left(0, \frac{3}{2}\right)=-\frac{27}{4}\) The portion of the graph between the dots corresponds to the solution curve satisfying the intial condition. To determine the interval of definition we find \(d y / d x\) for \\[2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\\]. Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\) and using a CAS to solve for \(x\) we get \(x=-1.13232 .\) Similarly, letting \(y=2,\) we find \(x=1.71299 .\) The largest interval of definition is approximately (-1.13232,1.71299). (c) The value of \(c\) corresponding to \(y(0)=-2\) is \(f(0,-2)=-40\) The portion of the graph to the right of the dot corresponds to the solution curve satisfying the initial condition. To determine the interval of definition we find \(d y / d x\) for \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\). Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\) and using a CAS to solve for \(x\) we get \(x=-2.29551 .\) The largest interval of definition is approximately \((-2.29551, \infty)\).

Differentiating \(\ln \left(x^{2}+10\right)+\csc y=c\) we get \\[\begin{array}{c}\frac{2 x}{x^{2}+10}-\csc y \cot y \frac{d y}{d x}=0 \\\\\frac{2 x}{x^{2}+10}-\frac{1}{\sin y} \cdot \frac{\cos y}{\sin y} \frac{d y}{d x}=0\end{array}\\] or \\[2 x \sin ^{2} y d x-\left(x^{2}+10\right) \cos y d y=0\\] Writing the differential equation in the form \(\frac{d y}{d x}=\frac{2 x \sin ^{2} y}{\left(x^{2}+10\right) \cos y}\) we see that singular solutions occur when \(\sin ^{2} y=0,\) or \(y=k \pi,\) where \(k\) is an integer.

(a) The differential equation is \(d P / d t=P(5-P)-4 .\) Solving \(P(5-P)-4=0\) for \(P\) we obtain equilibrium solutions \(P=1\) and \(P=4 .\) The phase portrait is shown on the right and solution curves are shown in part (b). We see that for \(P_{0}>4\) and \(1
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