/*! 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 11 (a) We have \(y(0)=c_{1}+c_{2}=0... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) We have \(y(0)=c_{1}+c_{2}=0, y^{\prime \prime}(1)=c_{1} e+c_{2} e^{-1}=1\) so that \(c_{1}=e /\left(e^{2}-1\right)\) and \(c_{2}=-e /\left(e^{2}-1\right) .\) The solution is \(y=e\left(e^{x}-e^{-x}\right) /\left(e^{2}-1\right)\) (b) We have \(y(0)=c_{3} \cosh 0+c_{4} \sinh 0=c_{3}=0\) and \(y(1)=c_{3} \cosh 1+c_{4} \sinh 1=c_{4} \sinh 1=1,\) so \(c_{3}=0\) and \(c_{4}=1 / \sinh 1 .\) The solution is \(y=(\sinh x) /(\sinh 1)\) (c) Starting with the solution in part (b) we have $$y=\frac{1}{\sinh 1} \sinh x=\frac{2}{e^{1}-e^{-1}} \frac{e^{x}-e^{-x}}{2}=\frac{e^{x}-e^{-x}}{e-1 / e}=\frac{e}{e^{2}-1}\left(e^{x}-e^{-x}\right).$$

Short Answer

Expert verified
The solutions for parts (a) and (c) are equivalent. Both are \(y = \frac{e}{e^2 - 1}(e^x - e^{-x})\).

Step by step solution

01

Verify Initial Conditions in (a)

The problem states that for part (a), the initial conditions are \(y(0) = c_1 + c_2 = 0\) and \(y''(1) = c_1 e + c_2 e^{-1} = 1\). Given \(c_1 = \frac{e}{e^2 - 1}\) and \(c_2 = -\frac{e}{e^2 - 1}\), substitute these values back to verify the conditions.For \(y(0)\): \[c_1 + c_2 = \frac{e}{e^2 - 1} + \left(-\frac{e}{e^2 - 1}\right) = 0\]For \(y''(1)\): \[c_1 e + c_2 e^{-1} = \frac{e}{e^2 - 1} e + \left(-\frac{e}{e^2 -1}\right) e^{-1} = 1\]Both conditions hold, verifying that the constants \(c_1\) and \(c_2\) are correctly defined.
02

Solution for (a)

Given \(c_1\) and \(c_2\), the solution for \(y\) in part (a) is:\[y = c_1 e^x + c_2 e^{-x} = \frac{e}{e^2-1}(e^x - e^{-x})\]This shows that the derived solution \(y = \frac{e(e^x - e^{-x})}{e^2 - 1}\) is correct.
03

Verify Initial Conditions in (b)

The problem states that for part (b), the initial conditions are \(y(0) = c_3 \cosh 0 + c_4 \sinh 0\) simplifies to \(c_3 = 0\) and \(y(1) = c_3 \cosh 1 + c_4 \sinh 1 = 1\) simplifies to \(c_4 \sinh 1 = 1\), giving \(c_4 = \frac{1}{\sinh 1}\).Both \(c_3\) and \(c_4\) satisfy the initial conditions.
04

Solution for (b)

With \(c_3 = 0\) and \(c_4 = \frac{1}{\sinh 1}\), the solution for \(y\) in part (b) is:\[y = c_3 \cosh x + c_4 \sinh x = \frac{\sinh x}{\sinh 1}\]
05

Rewrite Solution from (b) in Terms of Exponentials

The solution from (b), \(y = \frac{\sinh x}{\sinh 1}\), can be expressed using exponentials:\[y = \frac{1}{\sinh 1} \cdot \frac{e^x - e^{-x}}{2}\ = \frac{e^x - e^{-x}}{e - 1/e}\ = \frac{e}{e^2 - 1}(e^x - e^{-x})\]
06

Verify Equivalence of Solutions from (a) and (c)

Thus, we have rewritten and verified that the solution from part (c) matches the form of the solution from part (a), \(y = \frac{e}{e^2 - 1}(e^x - e^{-x})\). Both solutions are equivalent.

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Key Concepts

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

Initial Conditions
Initial conditions in differential equations are specific values given for a solution at a particular point, usually to find the constants of integration in a general solution. By specifying particular values at the start of a problem, often denoted as certain points along a function or its derivatives, these conditions help determine a unique solution to a differential equation. In the exercise provided, initial conditions play a crucial role.
  • For part (a) of the exercise, the initial conditions are given as \(y(0) = c_1 + c_2 = 0\) and \(y''(1) = c_1 e + c_2 e^{-1} = 1\).
  • These conditions allow us to determine the constants \(c_1\) and \(c_2\) necessary for a unique solution.
In solving these, we observe that using the specified initial conditions guides us in solving the coefficients precisely, resulting in solutions that meet the original problem's constraints. It is this process which distinguishes unique solutions in a landscape of possibilities.
Exponential Functions
Exponential functions are a foundational element in solving many differential equations, especially those reflecting growth and decay models. They are expressed in the form \(y = e^{kx}\), where the base \(e\) is the natural exponential constant, approximately equal to 2.71828.
  • In the context of part (a) and (c) of the exercise, we use exponentials to express the solution in terms of \(e^x\) and \(e^{-x}\).
  • This use of exponential functions leverages their property of transforming linear differential equations into algebraic ones, which are easier to solve.
By substituting exponential expressions back into the differential equation, we can verify the correctness of potential solutions. In our example, it helps demonstrate the correspondence and equivalence between solutions derived in different parts of the exercise, providing a consistent approach across varied mathematical problems.
Hyperbolic Functions
Hyperbolic functions, which include the hyperbolic sine (\(\sinh\)) and cosine (\(\cosh\)), are another form of mathematical expressions that often appear in solutions to differential equations, particularly those involving reflections similar to exponential behavior. They are defined as combinations of exponential functions:
  • \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
  • \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)
In part (b) of the exercise, we see the solution expressed using hyperbolic functions: \(y = \frac{\sinh x}{\sinh 1}\).It demonstrates how hyperbolic functions provide a compact and efficient way to express complex solutions. These functions share similarities with trigonometric functions, which makes them particularly useful in equations that reflect circular or periodic phenomena, although they pertain more to hyperbolic structures. By understanding their definitions and properties, we can navigate through differential equations with greater ease, leveraging known identities to check and simplify solutions.

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Most popular questions from this chapter

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 \(\sin \alpha x\)

Let \((x, y)\) be the coordinates of \(S_{2}\) on the curve \(C .\) The slope at \((x, y)\) is then \\[ d y / d x=\left(v_{1} t-y\right) /(0-x)=\left(y-v_{1} t\right) / x \\] or \(x y^{\prime}-y=-v_{1} t\) Differentiating with respect to \(x\) and using \(r=v_{1} / v_{2}\) gives \\[ \begin{aligned} x y^{\prime \prime}+y^{\prime}-y^{\prime} &=-v_{1} \frac{d t}{d x} \\ x y^{\prime \prime} &=-v_{1} \frac{d t}{d s} \frac{d s}{d x} \\ x y^{\prime \prime} &=-v_{1} \frac{1}{v_{2}}(-\sqrt{1+\left(y^{\prime}\right)^{2}}) \\ x y^{\prime \prime} &=r \sqrt{1+\left(y^{\prime}\right)^{2}} \end{aligned} \\] Letting \(u=y^{\prime}\) and separating variables, we obtain \\[ \begin{aligned} x \frac{d u}{d x} &=r \sqrt{1+u^{2}} \\ \frac{d u}{\sqrt{1+u^{2}}} &=\frac{r}{x} d x \\ \sinh ^{-1} u &=r \ln x+\ln c=\ln \left(c x^{r}\right) \\ u &=\sinh \left(\ln c x^{r}\right) \\ \frac{d y}{d x} &=\frac{1}{2}\left(c x^{r}-\frac{1}{c x^{r}}\right) \end{aligned} \\] At \(t=0, d y / d x=0\) and \(x=a,\) so \(0=c a^{r}-1 / c a^{r} .\) Thus \(c=1 / a^{r}\) and \\[ \frac{d y}{d x}=\frac{1}{2}\left[\left(\frac{x}{a}\right)^{r}-\left(\frac{a}{x}\right)^{r}\right]=\frac{1}{2}\left[\left(\frac{x}{a}\right)^{r}-\left(\frac{x}{a}\right)^{-r}\right] \\] If \(r>1\) or \(r<1,\) integrating gives \\[ y=\frac{a}{2}\left[\frac{1}{1+r}\left(\frac{x}{a}\right)^{1+r}-\frac{1}{1-r}\left(\frac{x}{a}\right)^{1-r}\right]+c_{1} \\] When \(t=0, y=0\) and \(x=a,\) so \(0=(a / 2)[1 /(1+r)-1 /(1-r)]+c_{1} .\) Thus \(c_{1}=a r /\left(1-r^{2}\right)\) and \\[ y=\frac{a}{2}\left[\frac{1}{1+r}\left(\frac{x}{a}\right)^{1+r}-\frac{1}{1-r}\left(\frac{x}{a}\right)^{1-r}\right]+\frac{a r}{1-r^{2}} \\] To see if the paths ever intersect we first note that if \(r>1\), then \(v_{1}>v_{2}\) and \(y \rightarrow \infty\) as \(x \rightarrow 0^{+} .\) In other words, \(S_{2}\) always lags behind \(S_{1}\). Next, if \(r<1\), then \(v_{1}

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 $$y=u e^{x}=\frac{5}{2} e^{3 x}+c_{1} e^{2 x}+c_{2} e^{x}$$

(a) The general solution of the differential equation is \(y=c_{1} \cos 4 x+c_{2} \sin 4 x .\) From \(y_{0}=y(0)=c_{1}\) we see that \(y=y_{0} \cos 4 x+c_{2} \sin 4 x .\) From \(y_{1}=y(\pi / 2)=y_{0}\) we see that any solution must satisfy \(y_{0}=y_{1} .\) We also see that when \(y_{0}=y_{1}, y=y_{0} \cos 4 x+c_{2} \sin 4 x\) is a solution of the boundary-value problem for any choice of \(c_{2}\). Thus, the boundary-value problem does not have a unique solution for any choice of \(y_{0}\) and \(y_{1}\). (b) Whenever \(y_{0}=y_{1}\) there are infinitely many solutions. (c) When \(y_{0} \neq y_{1}\) there will be no solutions. (d) The boundary-value problem will have the trivial solution when \(y_{0}=y_{1}=0 .\) This solution will not be unique.

Solving \(\frac{1}{20} q^{\prime \prime}+2 q^{\prime}+100 q=0\) we obtain \(q(t)=e^{-20 t}\left(c_{1} \cos 40 t+c_{2} \sin 40 t\right) .\) The initial conditions \(q(0)=5\) and \(q^{\prime}(0)=0\) imply \(c_{1}=5\) and \(c_{2}=5 / 2 .\) Thus \\[q(t)=e^{-20 t}\left(5 \cos 40 t+\frac{5}{2} \sin 40 t\right)=\sqrt{25+25 / 4} e^{-20 t} \sin (40 t+1.1071)\\] and \(q(0.01) \approx 4.5676\) coulombs. The charge is zero for the first time when \(40 t+1.1071=\pi\) or \(t \approx\) 0.0509 second.

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