/*! 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 19 Define \(y=u(x) e^{x}\) so $$y... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 19

Define \(y=u(x) e^{x}\) so $$y^{\prime}=u e^{x}+u^{\prime} e^{x}, \quad y^{\prime \prime}=u^{\prime \prime} e^{x}+2 u^{\prime} e^{x}+u e^{x}$$ and $$y^{\prime \prime}-3 y^{\prime}+2 y=e^{x} u^{\prime \prime}-e^{x} u^{\prime}=5 e^{3 x}$$ If \(w=u^{\prime}\) we obtain the linear first-order equation \(w^{\prime}-w=5 e^{2 x}\) which has the integrating factor \(e^{-\int d x}=e^{-x}\) Now $$\frac{d}{d x}\left[e^{-x} w\right]=5 e^{x} \quad \text { gives } \quad e^{-x} w=5 e^{x}+c_{1}$$ Therefore \(w=u^{\prime}=5 e^{2 x}+c_{1} e^{x}\) and \(u=\frac{5}{2} e^{2 x}+c_{1} e^{x}+c_{2} .\) The general solution is $$y=u e^{x}=\frac{5}{2} e^{3 x}+c_{1} e^{2 x}+c_{2} e^{x}$$

Short Answer

Expert verified
The general solution is \( y = \frac{5}{2} e^{3x} + c_1 e^{2x} + c_2 e^x \).

Step by step solution

01

Define the Function and Differentiate

We are given the function \( y = u(x) e^x \). To find the first derivative \( y' \), use the product rule: \( y' = u(x)' e^x + u(x) e^x = u' e^x + u e^x \).
02

Differentiate Again for Second Derivative

Differentiate \( y' = u' e^x + u e^x \) to find \( y'' \). Again applying the product rule, \( y'' = (u'' e^x + u' e^x) + (u' e^x + u e^x) = u'' e^x + 2u' e^x + u e^x \).
03

Substitute in Given Equation

Substitute \( y'' \) and \( y' \) into the equation given: \( y'' - 3y' + 2y = u'' e^x + 2u' e^x + u e^x - 3(u' e^x + u e^x) + 2(u e^x) \).
04

Simplify the Equation

Simplify the equation: \( y'' - 3y' + 2y = u'' e^x - u' e^x = 5e^{3x} \). Thus, \( e^x (u'' - u') = 5e^{3x} \).
05

Substitute and Simplify Further

Use the substitution \( w = u' \), giving \( w' = u'' \). So, \( w' - w = 5e^{2x} \).
06

Solve the Differential Equation

The differential equation is \( w' - w = 5e^{2x} \). Use an integrating factor \( e^{-x} \). Multiply through by this factor to obtain \( \frac{d}{dx}(e^{-x}w) = 5 e^{x} \).
07

Integrate and Solve for \( w \)

Integrate both sides: \( e^{-x}w = 5 \int e^{x} \, dx = 5 e^{x} + c_1 \). So, \( w = u' = 5e^{2x} + c_1 e^x \).
08

Integrate to Find \( u \)

Integrate \( u' = 5e^{2x} + c_1 e^x \) to find \( u \). This gives \( u = \frac{5}{2}e^{2x} + c_1 e^x + c_2 \).
09

Find the General Solution for \( y \)

Substitute \( u(x) \) back into the expression for \( y \): \( y = u e^x = \left( \frac{5}{2} e^{2x} + c_1 e^x + c_2 \right) e^x = \frac{5}{2} e^{3x} + c_1 e^{2x} + c_2 e^x \).

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Key Concepts

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

Product Rule
The product rule is a fundamental concept in calculus used when differentiating products of functions. It states that if you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product \( y = u(x)v(x) \) is given by:
  • \( y' = u'(x)v(x) + u(x)v'(x) \)
This rule helps break down the process of differentiating more complex expressions. For example, in our exercise, \( y = u(x)e^x \), and by applying the product rule, we find:
  • \( y' = u'(x)e^x + u(x)e^x \)
Here, each function is differentiated separately, making the computation much faster and simpler. Always keep in mind that without the product rule, attempting to differentiate such equations directly becomes challenging.
Integrating Factor
The concept of an integrating factor is used to solve linear first-order differential equations, which take the form \( y' + P(x)y = Q(x) \). An integrating factor is a function \( \mu(x) \) such that multiplication converts the differential equation into an easier-to-solve form. In this exercise, for the equation \( w' - w = 5e^{2x} \), the integrating factor is \( e^{-x} \). This factor arises because we aim to rewrite the differential as:
  • \( \frac{d}{dx}(e^{-x}w) = 5e^x \)
Multiplying through by the integrating factor helps to consolidate terms and establish a straightforward integration path. Once the left-hand side is expressed as a derivative of a single function, integrating both sides provides an answer directly related to the problem.
First-Order Differential Equation
A first-order differential equation involves the derivative of the function you鈥檙e solving for but does not involve higher-order derivatives. Equations like \( w' - w = 5e^{2x} \) are classical examples of first-order linear differential equations. Their solutions depict the rate of change for a particular function, often correlating to real-world scenarios such as population growth or decay. In our exercise, transforming the problem through functions \( u \) and \( w \), we reduce complexity and directly deal with the primary differentials needed for solving.
General Solution
The general solution is the function form that includes all possible solutions of a differential equation considering arbitrary constants that arise during integration. These constants are often determined by initial conditions or boundary values provided within specific problems. In our previous steps, integrating and solving gives us:
  • \( w = u' = 5e^{2x} + c_1 e^x \)
  • \( u = \frac{5}{2}e^{2x} + c_1 e^x + c_2 \)
Continuing to find \( y \), we reach:
  • \( y = \frac{5}{2} e^{3x} + c_1 e^{2x} + c_2 e^x \)
This solution accounts for the different possible forms \( the solution \) might adopt depending on specific circumstances defined by \( c_1 \) and \( c_2 \), therefore covering a broad set of scenarios.

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Most popular questions from this chapter

(a) Setting \(d x / d t=v,\) the differential equation becomes \((L-x) d v / d t-v^{2}=L g .\) But, by the Chain Rule, \(d v / d t=(d v / d x)(d x / d t)=v d v / d x,\) so \((L-x) v d v / d x-v^{2}=L g .\) Separating variables and integrating we obtain \(\frac{v}{v^{2}+L g} d v=\frac{1}{L-x} d x \quad\) and \(\quad \frac{1}{2} \ln \left(v^{2}+L g\right)=-\ln (L-x)+\ln c\) so \(\sqrt{v^{2}+L g}=c /(L-x) .\) When \(x=0, v=0,\) and \(c=L \sqrt{L g} .\) Solving for \(v\) and simplifying we get \\[ \frac{d x}{d t}=v(x)=\frac{\sqrt{L g\left(2 L x-x^{2}\right)}}{L-x} \\] Again, separating variables and integrating we obtain \\[ \frac{L-x}{\sqrt{L g\left(2 L x-x^{2}\right)}} d x=d t \quad \text { and } \quad \frac{\sqrt{2 L x-x^{2}}}{\sqrt{L g}}=t+c_{1} \\] since \(x(0)=0,\) we have \(c_{1}=0\) and \(\sqrt{2 L x-x^{2}} / \sqrt{L g}=t .\) Solving for \(x\) we get \\[ x(t)=L-\sqrt{L^{2}-L g t^{2}} \quad \text { and } \quad v(t)=\frac{d x}{d t}=\frac{\sqrt{L} g t}{\sqrt{L-g t^{2}}} \\] (b) The chain will be completely on the ground when \(x(t)=L\) or \(t=\sqrt{L / g}\) (c) The predicted velocity of the upper end of the chain when it hits the ground is infinity.

Using a CAS to solve the auxiliary equation \(m^{4}+2 m^{2}-m+2=0\) we find \(m_{1}=1 / 2+\sqrt{3} i / 2, m_{2}=1 / 2-\sqrt{3} i / 2\) \(m_{3}=-1 / 2+\sqrt{7} i / 2,\) and \(m_{4}=-1 / 2-\sqrt{7} i / 2 .\) The general solution is $$y=e^{x / 2}\left(c_{1} \cos \frac{\sqrt{3}}{2} x+c_{2} \sin \frac{\sqrt{3}}{2} x\right)+e^{-x / 2}\left(c_{3} \cos \frac{\sqrt{7}}{2} x+c_{4} \sin \frac{\sqrt{7}}{2} x\right)$$

If \(1-i\) is a root of the auxiliary equation then so is \(1+i,\) and the auxiliary equation is $$(m-2)[m-(1+i)][m-(1-i)]=m^{3}-4 m^{2}+6 m-4=0.$$ We need \(m^{3}-4 m^{2}+6 m-4\) to have the form \(m(m-1)(m-2)+b m(m-1)+c m+d .\) Expanding this last expression and equating coefficients we get \(b=-1, c=3,\) and \(d=-4 .\) Thus, the differential equation is $$x^{3} y^{\prime \prime \prime}-x^{2} y^{\prime \prime}+3 x y^{\prime}-4 y=0.$$

The solution of the initial-value problem \\[ x^{\prime \prime}+\omega^{2} x=0, \quad x(0)=0, x^{\prime}(0)=v_{0}, \omega^{2}=10 / m \\] is \(x(t)=\left(v_{0} / \omega\right) \sin \omega t .\) To satisfy the additional boundary condition \(x(1)=0\) we require that \(\omega=n \pi\) \(n=1,2,3, \ldots,\) The eigenvalues \(\lambda=\omega^{2}=n^{2} \pi^{2}\) and eigenfunctions of the problem are then \(x(t)=\) \(\left(v_{0} / n \pi\right) \sin n \pi t . \quad\) Using \(\omega^{2}=10 / m\) we find that the only masses that can pass through the equilibrium position at \(t=1\) are \(m_{n}=10 / n^{2} \pi^{2} .\) Note for \(n=1,\) the heaviest mass \(m_{1}=10 / \pi^{2}\) will not pass through the equilibrium position on the interval \(0 < t<1\) (the period of \(x(t)=\left(v_{0} / \pi\right) \sin \pi t\) is \(T=2,\) so on \(0 \leq t \leq 1\) its graph passes through \(x=0\) only at \(t=0\) and \(t=1\) ). Whereas for \(n>1\), masses of lighter weight will pass through the equilibrium position \(n-1\) times prior to passing through at \(t=1 .\) For example, if \(n=2,\) the period of \(x(t)=\left(v_{0} / 2 \pi\right) \sin 2 \pi t\) is \(2 \pi / 2 \pi=1,\) the mass will pass through \(x=0\) only once \(\left(t=\frac{1}{2}\right)\) prior to \(t=1 ;\) if \(n=3,\) the period of \(x(t)=\left(v_{0} / 3 \pi\right) \sin 3 \pi t\) is \(\frac{2}{3},\) the mass will pass through \(x=0\) twice \(\left(t=\frac{1}{3} \text { and } t=\frac{2}{3}\right)\) prior to \(t=1 ;\) and so on.

From Newton's second law in the \(x\) -direction we have $$m \frac{d^{2} x}{d t^{2}}=-k \cos \theta=-k \frac{1}{v} \frac{d x}{d t}=-|c| \frac{d x}{d t}.$$ In the \(y\) -direction we have $$m \frac{d^{2} y}{d t^{2}}=-m g-k \sin \theta=-m g-k \frac{1}{v} \frac{d y}{d t}=-m g-|c| \frac{d y}{d t}$$ From \(m D^{2} x+|c| D x=0\) we have \(D(m D+|c|) x=0\) so that \((m D+|c|) x=c_{1}\) or \((D+|c| / m) x=c_{2} .\) This is a linear first-order differential equation. An integrating factor is \(e^{\int|c| d t / m}=e^{|c| t / m}\) so that $$\frac{d}{d t}\left[e^{|c| t / m} x\right]=c_{2} e^{|c| t / m}$$ and \(e^{|c| t / m} x=\left(c_{2} m /|c|\right) e^{|c| t / m}+c_{3} .\) The general solution of this equation is \(x(t)=c_{4}+c_{3} e^{-|c| t / m} .\) From \(\left(m D^{2}+|c| D\right) y=-m g\) we have \(D(m D+|c|) y=-m g\) so that \((m D+|c|) y=-m g t+c_{1}\) or \((D+|c| / m) y=-g t+c_{2}\) This is a linear first-order differential equation with integrating factor \(e^{\int|c| d t / m}=e^{|c| t / m} .\) Thus $$\begin{aligned} \frac{d}{d t}\left[e^{|c| t / m} y\right] &=\left(-g t+c_{2}\right) e^{|c| t / m} \\ e^{|c| t / m} y &=-\frac{m g}{|c|} t e^{|c| t / m}+\frac{m^{2} g}{c^{2}} e^{|c| t / m}+c_{3} e^{|c| t / m}+c_{4} \end{aligned}$$ and $$y(t)=-\frac{m g}{|c|} t+\frac{m^{2} g}{c^{2}}+c_{3}+c_{4} e^{-|c| t / m}.$$
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