/*! 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 16 The thinner curve is obtained us... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 16

The thinner curve is obtained using a numerical solver, while the thicker curve is the graph of the Taylor polynomial. We look for a solution of the form \\[ \begin{array}{c} y(x)=y(0)+y^{\prime}(0) x+\frac{1}{2 !} y^{\prime \prime}(0) x^{2}+\frac{1}{3 !} y^{\prime \prime \prime}(0) x^{3}+\frac{1}{4 !} y^{(4)}(0) x^{4} \\ +\frac{1}{5 !} y^{(5)}(0) x^{5}+\frac{1}{6 !} y^{(6)}(0) x^{6} \end{array} \\] From \(y^{\prime \prime}(x)=e^{y}\) we compute \\[ \begin{aligned} y^{\prime \prime \prime}(x) &=e^{y} y^{\prime} \\ y^{(4)}(x) &=e^{y}\left(y^{\prime}\right)^{2}+e^{y} y^{\prime \prime} \\ y^{(5)}(x) &=e^{y}\left(y^{\prime}\right)^{3}+3 e^{y} y^{\prime} y^{\prime \prime}+e^{y} y^{\prime \prime \prime} \\ y^{(6)}(x) &=e^{y}\left(y^{\prime}\right)^{4}+6 e^{y}\left(y^{\prime}\right)^{2} y^{\prime \prime}+3 e^{y}\left(y^{\prime \prime}\right)^{2}+4 e^{y} y^{\prime} y^{\prime \prime \prime}+e^{y} y^{(4)} \end{aligned} \\] Using \(y(0)=0\) and \(y^{\prime}(0)=-1\) we find \\[ y^{\prime \prime}(0)=1, \quad y^{\prime \prime \prime}(0)=-1, \quad y^{(4)}(0)=2, \quad y^{(5)}(0)=-5, \quad y^{(6)}(0)=16 \\] An approximate solution is \\[ y(x)=-x+\frac{1}{2} x^{2}-\frac{1}{6} x^{3}+\frac{1}{12} x^{4}+\frac{1}{24} x^{5}+\frac{1}{45} x^{6} .\\]

Short Answer

Expert verified
The approximate solution is \(y(x) = -x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{12}x^4 - \frac{1}{24}x^5 + \frac{1}{45}x^6\).

Step by step solution

01

Identify Given Information

The exercise involves finding an approximate solution to a differential equation using a Taylor polynomial. You are given the initial values: \(y(0) = 0\), \(y'(0) = -1\), and derived values such as \(y''(0) = 1\), \(y'''(0) = -1\), and so on.
02

Write the General Taylor Series

The general form of the Taylor series given is: \[ y(x) = y(0) + y'(0)x + \frac{1}{2!}y''(0)x^2 + \frac{1}{3!}y'''(0)x^3 + \frac{1}{4!}y^{(4)}(0)x^4 + \frac{1}{5!}y^{(5)}(0)x^5 + \frac{1}{6!}y^{(6)}(0)x^6 \].This formula represents the sum of derivatives evaluated at zero multiplied by increasing powers of \(x\) with corresponding factorial denominators.
03

Substitute the Derivative Values

Plug in the given derivatives at \(x = 0\) into the Taylor series formula:\[ y(x) = 0 + (-1)x + \frac{1}{2!}(1)x^2 + \frac{1}{3!}(-1)x^3 + \frac{1}{4!}(2)x^4 + \frac{1}{5!}(-5)x^5 + \frac{1}{6!}(16)x^6 \].
04

Simplify Each Term

Perform the necessary multiplications and simplifications for each term:- The first term is \(-x\).- The second term is \(\frac{1}{2}x^2\).- The third term is \(-\frac{1}{6}x^3\).- The fourth term is \(\frac{1}{12}x^4\).- The fifth term is \(-\frac{1}{24}x^5\).- The sixth term is \(\frac{1}{45}x^6\).
05

Write the Approximate Solution

The final approximate solution for \(y(x)\) is therefore:\[ y(x) = -x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{12}x^4 - \frac{1}{24}x^5 + \frac{1}{45}x^6 \].

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Key Concepts

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

Numerical Solver
A numerical solver is a computational tool used to approximate the solutions of mathematical problems, particularly when an exact solution cannot be easily obtained. In differential equations, this tool is highly beneficial because these equations often describe how quantities change with respect to one another, and precise analytical solutions can be complex or unavailable. Numerical solvers employ algorithms that iteratively perform calculations to reach an approximate answer.
- They are crucial for solving high-order differential equations - Used heavily in scientific computing and programming
Numerical solvers can work alongside Taylor series expansions to provide reliable approximations of functions. While the Taylor polynomial gives a local approximation around a certain point, numerical solvers can help explore these functions over broader domains, ensuring that predictions remain consistent with the underlying model.
Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. They are widely used in various fields to model the behavior of dynamic systems. In the given exercise, the differential equation of interest is second order: \(y''(x) = e^y\)
This signifies that the second derivative of \(y\) with respect to \(x\) is equal to the exponential of \(y\).
- They can be ordinary (ODEs) or partial (PDEs) based on the number of variables involved- Describe the change and movement of phenomena like heat, sound, electricity, etc.
For our problem, solving this differential equation directly may not be straightforward due to the nonlinear term \(e^y\). Thus, using a Taylor polynomial offers a way to approximately solve the differential equation in the vicinity of a starting point. Determining initial conditions like \(y(0) = 0\) and \(y'(0) = -1\) helps to uniquely identify the specific solution needed for further analysis.
Series Expansion
Series expansion represents a function as a sum of terms derived from the function's derivatives at a particular point. The Taylor series is one of the most common forms of series expansions. This provides a polynomial that approximates a given function in its neighborhood around a center point, often zero.
- The Taylor series is crucial for approximating functions that are otherwise challenging to express- It uses derivatives to construct an infinite sequence of terms that, in ideal cases, converge to the exact function
Each term in the Taylor polynomial for the problem represents the function's behavior around zero, using derivatives evaluated at zero:\[ y(x) = y(0) + y'(0)x + \frac{1}{2!}y''(0)x^2 + \dots \]
The approximation becomes more accurate with more terms. However, practical computation often limits the expansion to a finite number of terms, expanding until \(x^6\) in the given exercise. Such expansion helps in approximating solutions of differential equations when analytical solutions are hard to determine.

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Most popular questions from this chapter

The auxiliary equation is \(m^{2}=0\) so that \(u(r)=c_{1}+c_{2} \ln r .\) The boundary conditions \(u(a)=u_{0}\) and \(u(b)=u_{1}\) yield the system \(c_{1}+c_{2} \ln a=u_{0}, c_{1}+c_{2} \ln b=u_{1} .\) Solving gives \\[ c_{1}=\frac{u_{1} \ln a-u_{0} \ln b}{\ln (a / b)} \quad \text { and } \quad c_{2}=\frac{u_{0}-u_{1}}{\ln (a / b)} \\]. Thus \\[ u(r)=\frac{u_{1} \ln a-u_{0} \ln b}{\ln (a / b)}+\frac{u_{0}-u_{1}}{\ln (a / b)} \ln r=\frac{u_{0} \ln (r / b)-u_{1} \ln (r / a)}{\ln (a / b)} \\].

From \(\frac{5}{16} x^{\prime \prime}+\beta x^{\prime}+5 x=0\) we find that the roots of the auxiliary equation are \(m=-\frac{8}{5} \beta \pm \frac{4}{5} \sqrt{4 \beta^{2}-25}\) (a) If \(4 \beta^{2}-25>0\) then \(\beta>5 / 2\) (b) If \(4 \beta^{2}-25=0\) then \(\beta=5 / 2\) (c) If \(4 \beta^{2}-25<0\) then \(0<\beta<5 / 2\).

For \(\lambda=\alpha^{4}, \alpha>0,\) the general solution of the boundary-value problem \\[ y^{(4)}-\lambda y=0, \quad y(0)=0, y^{\prime \prime}(0)=0, y(1)=0, y^{\prime \prime}(1)=0 \\] is \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x+c_{3} \cosh \alpha x+c_{4} \sinh \alpha x\). The boundary conditions \(y(0)=0, y^{\prime \prime}(0)=0\) give \(c_{1}+c_{3}=0\) and \(-c_{1} \alpha^{2}+c_{3} \alpha^{2}=0,\) from which we conclude \(c_{1}=c_{3}=0 .\) Thus, \(y=c_{2} \sin \alpha x+c_{4} \sinh \alpha x .\) The boundary conditions \(y(1)=0, y^{\prime \prime}(1)=0\) then give \\[ \begin{aligned} c_{2} \sin \alpha+c_{4} \sinh \alpha &=0 \\ -c_{2} \alpha^{2} \sin \alpha+c_{4} \alpha^{2} \sinh \alpha &=0 \end{aligned} \\]. In order to have nonzero solutions of this system, we must have the determinant of the coefficients equal zero, that is, \\[ \left|\begin{array}{cc} \sin \alpha & \sinh \alpha \\ -\alpha^{2} \sin \alpha & \alpha^{2} \sinh \alpha \end{array}\right|=0 \quad \text { or } \quad 2 \alpha^{2} \sinh \alpha \sin \alpha=0 \\] But since \(\alpha>0,\) the only way that this is satisfied is to have \(\sin \alpha=0\) or \(\alpha=n \pi .\) The system is then satisfied by choosing \(c_{2} \neq 0, c_{4}=0,\) and \(\alpha=n \pi .\) The eigenvalues and corresponding eigenfunctions are then \\[ \lambda_{n}=\alpha^{4}=(n \pi)^{4}, n=1,2,3, \ldots \quad \text { and } \quad y=\sin n \pi x \\].

The thinner curve is obtained using a numerical solver, while the thicker curve is the graph of the Taylor polynomial. We look for a solution of the form \\[ y(x)=y(0)+y^{\prime}(0) x+\frac{1}{2 !} y^{\prime \prime}(0) x^{2}+\frac{1}{3 !} y^{\prime \prime \prime}(0) x^{3}+\frac{1}{4 !} y^{(4)}(0) x^{4}+\frac{1}{5 !} y^{(5)}(0) x^{5} \\] From \(y^{\prime \prime}(x)=1-y^{2}\) we compute \\[ \begin{aligned} y^{\prime \prime \prime}(x) &=-2 y y^{\prime} \\ y^{(4)}(x) &=-2 y y^{\prime \prime}-2\left(y^{\prime}\right)^{2} \\ y^{(5)}(x) &=-2 y y^{\prime \prime \prime}-6 y^{\prime} y^{\prime \prime} \end{aligned} \\] Using \(y(0)=2\) and \(y^{\prime}(0)=3\) we find \\[ y^{\prime \prime}(0)=-3, \quad y^{\prime \prime \prime}(0)=-12, \quad y^{(4)}(0)=-6, \quad y^{(5)}(0)=102 \\] An approximate solution is \\[ y(x)=2+3 x-\frac{3}{2} x^{2}-2 x^{3}-\frac{1}{4} x^{4}+\frac{17}{20} x^{5} \\]

Applying integration by parts twice we have $$\begin{aligned} \int e^{a x} f(x) d x &=\frac{1}{a} e^{a x} f(x)-\frac{1}{a} \int e^{a x} f^{\prime}(x) d x \\ &=\frac{1}{a} e^{a x} f(x)-\frac{1}{a}\left[\frac{1}{a} e^{a x} f^{\prime}(x)-\frac{1}{a} \int e^{a x} f^{\prime \prime}(x) d x\right] \\ &=\frac{1}{a} e^{a x} f(x)-\frac{1}{a^{2}} e^{a x} f^{\prime}(x)+\frac{1}{a^{2}} \int e^{a x} f^{\prime \prime}(x) d x \end{aligned}$$ Collecting the integrals we get $$\int e^{a x}\left(f(x)-\frac{1}{a^{2}} f^{\prime \prime}(x)\right) d x=\frac{1}{a} e^{a x} f(x)-\frac{1}{a^{2}} e^{a x} f^{\prime}(x)$$ In order for the technique to work we need to have $$\begin{aligned} &\int e^{a x}\left(f(x)-\frac{1}{a^{2}} f^{\prime \prime}(x)\right) d x=k \int e^{a x} f(x) d x\\\ &f(x)-\frac{1}{a^{2}} f^{\prime \prime}(x)=k f(x) \end{aligned}$$ where \(k \neq 0 .\) This is the second-order differential equation \(f^{\prime \prime}(x)+a^{2}(k-1) f(x)=0\) If \(k<1, k \neq 0,\) the solution of the differential equation is a pair of exponential functions, in which case the original integrand is an exponential function and does not require integration by parts for its evaluation. Similarly, if \(k=1, f^{\prime \prime}(x)=0\) and \(f(x)\) has the form \(f(x)=a x+b .\) In this case a single application of integration by parts will suffice. Finally, if \(k>1\), the solution of the differential equation is $$f(x)=c_{1} \cos a \sqrt{k-1} x+c_{2} \sin a \sqrt{k-1} x$$ and we see that the technique will work for linear combinations of \(\cos \alpha x\) and \(\sin \alpha x\)
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