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(a) The substitutions \(y=y_{1}+u\) and $$\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d u}{d x}$$ lead to $$\begin{aligned} \frac{d y_{1}}{d x}+\frac{d u}{d x} &=P+Q\left(y_{1}+u\right)+R\left(y_{1}+u\right)^{2} \\ &=P+Q y_{1}+R y_{1}^{2}+Q u+2 y_{1} R u+R u^{2} \end{aligned}$$ or $$\frac{d u}{d x}-\left(Q+2 y_{1} R\right) u=R u^{2}.$$ This is a Bernoulli equation with \(n=2\) which can be reduced to the linear equation $$\frac{d w}{d x}+\left(Q+2 y_{1} R\right) w=-R$$ by the substitution \(w=u^{-1}\) (b) Identify \(P(x)=-4 / x^{2}, Q(x)=-1 / x,\) and \(R(x)=1 .\) Then \(\frac{d w}{d x}+\left(-\frac{1}{x}+\frac{4}{x}\right) w=-1 .\) An integrating factor is \(x^{3}\) so that \(x^{3} w=-\frac{1}{4} x^{4}+c\) or \(u=\left[-\frac{1}{4} x+c x^{-3}\right]^{-1} .\) Thus, \(y=\frac{2}{x}+u\).

Short Answer

Expert verified
The solution yields \( y = \frac{2}{x} + u \), where \( u = \left[-\frac{1}{4}x + cx^{-3}\right]^{-1} \).

Step by step solution

01

Apply Substitution

We begin by substituting \( y = y_{1} + u \) into the original equation. This gives us two substitutions: \( y = y_{1} + u \) and \( \frac{d y}{d x} = \frac{d y_{1}}{d x} + \frac{d u}{d x} \). These substitutions will help simplify the differential equation.
02

Organize Terms

Substitute the expressions into the given equation: \( \frac{d y_{1}}{d x} + \frac{d u}{d x} = P + Q(y_{1}+u) + R(y_{1}+u)^2 \). Expanding and organizing these terms gives \( \frac{d y_{1}}{d x} + \frac{d u}{d x} = P + Qy_{1} + Ry_{1}^2 + Qu + 2y_{1}Ru + Ru^2 \).
03

Isolate du/dx

Rearrange the equation to focus on \( \frac{d u}{d x} \), yielding: \( \frac{d u}{d x} - (Q + 2y_{1}R)u = Ru^2 \). This is in the Bernoulli form with \( n = 2 \).
04

Substitute for Bernoulli Equation

Convert the equation to a linear form by substituting \( w = u^{-1} \). This gives \( \frac{d w}{d x} + (Q + 2y_{1}R)w = -R \).
05

Identify Functions

Identify the functions from the problem statement: \( P(x) = -\frac{4}{x^2} \), \( Q(x) = -\frac{1}{x} \), and \( R(x) = 1 \). Substitute these functions into the linear equation, resulting in \( \frac{d w}{d x} - \left(-\frac{1}{x} + \frac{4}{x}\right) w = -1 \).
06

Integrating Factor

Find the integrating factor, which is \( x^{3} \), to solve the linear differential equation. Multiply throughout by the integrating factor to get: \( x^3 \frac{d w}{d x} + x^2 w = -x^3 \).
07

Solve the Linear Equation

Integrating both sides after finding the integrating factor leads to \( x^3 w = -\frac{1}{4}x^4 + c \), where \( c \) is the constant of integration.
08

Back Substitution

Solve for \( u \) by back-substituting \( w = u^{-1} \), yielding \( u = \left[-\frac{1}{4}x + cx^{-3}\right]^{-1} \). Finally, substitute back to get \( y = \frac{2}{x} + u \).

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Key Concepts

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

Differential Equations
Differential equations are equations that involve an unknown function and its derivatives. They are used to describe various phenomena in engineering, physics, finance, and other fields. An equation is considered a differential equation if it includes derivatives of one or more dependent variables with respect to one or more independent variables.
For instance, in our exercise, we deal with the differential equation involving the derivatives of a function of one variable, typically expressed as \( \frac{dy}{dx} \). This represents how the function \( y \) changes as \( x \) changes, which is central to understanding how different quantities relate over time or space.
In a Bernoulli differential equation, a subset of nonlinear differential equations, the structure generally involves the term \( u^n \), where \( n \) is a constant. The equation can take the shape \( \frac{du}{dx} + a(x)u = b(x)u^n \). Such equations require specific methods to solve, such as substitution or transforming the equation to a different form, leading us into the substitution methods discussed in the next section.
Substitution Methods
Substitution methods are powerful techniques used to simplify differential equations, making them easier to solve. By using substitutions, we can transform a complex problem into a more manageable form.
In this exercise, we employ a substitution by letting \( y = y_1 + u \) and \( \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{du}{dx} \). This helps to separate the problem into components that highlight the nonlinear parts of the solution embeddable within the differential equation.
This substitution allows us to identify the underlying form of the Bernoulli equation \( \frac{du}{dx} - (Q + 2y_1R)u = Ru^2 \). By further substituting \( w = u^{-1} \), the equation transforms into a linear form, which is much simpler to solve. This step is critical for handling nonlinear equations, converting them to linear differential equations, for which there are established solving techniques.
Integrating Factor
The integrating factor is a crucial tool in solving linear differential equations. It transforms a non-separable linear equation into one that can be easily integrated. This is done by multiplying every term in the equation by this factor.
In our exercise, once the Bernoulli equation was transformed using a substitution, it took the form \( \frac{dw}{dx} + (Q + 2y_1R)w = -R \). The function that multiplies the derivative, here in our example is the sum \( (Q - 2y_1R) \), is used to find the integrating factor.
Specifically, for our transformed equation, the integrating factor was identified as \( x^3 \). We multiply the entire equation by \( x^3 \) to enable integration: \( x^3 \frac{dw}{dx} + x^2w = -x^3 \). After this, integrating both sides gives the solution in terms of \( w \), leading us finally to solve for \( u \), and then substituting back to find \( y \).
This process illustrates the power of integrating factors for solving differential equations, as they simplify the integration process and help in finding the general solution.

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Most popular questions from this chapter

(a) Solving \(v_{t}=\sqrt{m g / k}\) for \(k\) we obtain \(k=m g / v_{t}^{2} .\) The differential equation then becomes \\[ m \frac{d v}{d t}=m g-\frac{m g}{v_{t}^{2}} v^{2} \\] or \(\frac{d v}{d t}=g\left(1-\frac{1}{v_{t}^{2}} v^{2}\right)\) Separating variables and integrating gives \\[ v_{t} \tanh ^{-1} \frac{v}{v_{t}}=g t+c_{1} \\] The initial condition \(v(0)=0\) implies \(c_{1}=0,\) so \\[ v(t)=v_{t} \tanh \frac{g t}{v_{t}} \\] We find the distance by integrating: \\[ s(t)=\int v_{t} \tanh \frac{g t}{v_{t}} d t=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)+c_{2} \\] The initial condition \(s(0)=0\) implies \(c_{2}=0,\) so \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right) \\] In 25 seconds she has fallen \(20,000-14,800=5,200\) feet. Using a CAS to solve \\[ 5200=\left(v_{t}^{2} / 32\right) \ln \left(\cosh \frac{32(25)}{v_{t}}\right) \\] for \(v_{t}\) gives \(v_{t} \approx 271.711 \mathrm{ft} / \mathrm{s} .\) Then \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)=2307.08 \ln (\cosh 0.117772 t) \\] (b) \(\operatorname{At} t=15, s(15)=2,542.94 \mathrm{ft}\) and \(v(15)=s^{\prime}(15)=256.287 \mathrm{ft} / \mathrm{sec}.\)

(a) All solutions of the form \(y=x^{5} e^{x}-x^{4} e^{x}+c x^{4}\) satisfy the initial condition. In this case, since \(4 / x\) is discontinuous at \(x=0,\) the hypotheses of Theorem 1.1 are not satisfied and the initial-value problem does not have a unique solution. (b) The differential equation has no solution satisfying \(y(0)=y_{0}, y_{0}>0\) (c) In this case, since \(x_{0}>0,\) Theorem 1.1 applies and the initial-value problem has a unique solution given by \(y=x^{5} e^{x}-x^{4} e^{x}+c x^{4}\) where \(c=y_{0} / x_{0}^{4}-x_{0} e^{x_{0}}+e^{x_{0}}\)

(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 \((-2.29551, \infty)\).

We want the general solution to be \(y=3 x-5+c e^{-x}\). (Rather than \(e^{-x}\), any function that approaches 0 as \(x \rightarrow \infty\) could be used.) Differentiating we get \\[ y^{\prime}=3-c e^{-x}=3-(y-3 x+5)=-y+3 x-2 \\] so the differential equation \(y^{\prime}+y=3 x-2\) has solutions asymptotic to the line \(y=3 x-5\).

For \(y^{\prime}+\frac{1}{x+1} y=\frac{\ln x}{x+1}\) an integrating factor is \(e^{\int[1 /(x+1)] d x}=x+1\) so that \(\frac{d}{d x}[(x+1) y]=\ln x\) and \(y=\frac{x}{x+1} \ln x-\frac{x}{x+1}+\frac{c}{x+1}\) for \(0< x<\infty .\) If \(y(1)=10\) then \(c=21\) and \(y=\frac{x}{x+1} \ln x-\frac{x}{x+1}+\frac{21}{x+1}\).

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