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Chapter 3: Problem 3

Let \(u=y^{\prime}\) so that \(u^{\prime}=y^{\prime \prime} .\) The equation becomes \(u^{\prime}=-u-1\) which is separable. Thus \\[ \frac{d u}{u^{2}+1}=-d x \Longrightarrow \tan ^{-1} u=-x+c_{1} \Longrightarrow y^{\prime}=\tan \left(c_{1}-x\right) \Longrightarrow y=\ln \left|\cos \left(c_{1}-x\right)\right|+c_{2.} \\]

Short Answer

Expert verified
The solution is \( y = \ln \left| \cos \left(c_1 - x \right) \right| + c_2 \).

Step by step solution

01

Substitute and Differentiate

Recognize that by letting \( u = y' \), we are simplifying the second-order differential equation by reducing its order. The derivative \( u' \) then becomes equivalent to the second derivative of \( y \), written as \( y'' \). The given equation is \( u' = -u - 1 \).
02

Separate Variables

The differential equation \( u' = -u - 1 \) is separable, which means we can rearrange it to separate the variables. We can write it as \( \frac{du}{dx} = -u - 1 \), and then rearrange to isolate \( du \) and \( dx \): \( \frac{du}{u^2 + 1} = -dx \).
03

Integrate Both Sides

Integrate both sides of the equation: the left side with respect to \( u \) and the right side with respect to \( x \). The integral of \( \frac{du}{u^2 + 1} \) is \( \tan^{-1}(u) \), and the integral of \(-dx\) is \(-x + c_1\), where \(c_1\) is the constant of integration. Thus, \( \tan^{-1}(u) = -x + c_1 \).
04

Solve for \(u\)

To extract \(u = y'\), solve \( \tan^{-1}(u) = -x + c_1 \). This results in \( u = \tan(c_1 - x) \), meaning that \( y' = \tan(c_1 - x) \).
05

Integrate to Find \(y\)

Now integrate \( y' = \tan(c_1 - x) \) with respect to \( x \) to find \( y \). The integral of \( \tan(cx) \) is \( -\ln|\cos(cx)| \). Thus, \( y = \ln|\cos(c_1 - x)| + c_2 \), where \(c_2\) is another constant of integration.

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Key Concepts

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

Second-Order Differential Equations
Second-order differential equations are quite common in mathematical modeling, especially in physics and engineering. They are equations that involve the second derivative of a function, typically denoted as \( y'' \). In these problems, we often want to find a function \( y(x) \) that satisfies the equation involving its second derivative. Here, we've reduced the problem's order using a substitution method: letting \( u = y' \). By this choice, the original second-order problem is transformed into a first-order one, allowing easier solution strategies. This process simplifies our task because dealing with a first derivative (\( y' \)) is often less complicated than handling \( y'' \). Once reduced, you arrive at solving a first-order differential equation, making the overall process more straightforward. Understanding how to reduce the order of equations can significantly help tackle more complex differential equations.
Integration by Substitution
Integration by substitution is a powerful technique often used to solve integrals more easily. It involves changing the variable of integration to transform the integral into a simpler form that can be solved more directly.When you encounter a differential equation, like \( u' = -u - 1 \), which is separable, you can rearrange it to isolate the derivatives and variables. In our case, we rewrite it with \( \frac{du}{u^2 + 1} = -dx \), showing the separation. Then, substituting and integrating each side becomes straightforward. This technique transforms the integral into one that is recognizable and solvable with standard calculus methods. The substitution enables easier manipulation, which is critical in finding a solution.
Trigonometric Integration
Trigonometric integration involves using trigonometric identities to solve integrals that include trigonometric functions. Once the variables are separated and rearranged like we did with \( \frac{du}{u^2 + 1} = -dx \), a familiar form appears for integration. The integral \( \frac{du}{u^2 + 1} \) relates to \( \tan^{-1}(u) \), which is a standard form in trigonometric integration. Recognizing such forms helps in directly applying trigonometric identities or inverse functions, easing the integration process. Afterward, solving for \( u \) gives \( u = \tan(c_1 - x) \), leading to the integration of the tangent function. Lastly, the antiderivative of \( \tan(x) \) is \(-\ln|\cos(x)|\), derived using trigonometric integration techniques, letting us find \( y \). Trigonometric insights are very useful in solving complex integrals with trigonometric expressions.

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Most popular questions from this chapter

(a) A solution curve has the same \(y\) -coordinate at both ends of the interval \([-\pi, \pi]\) and the tangent lines at the endpoints of the interval are parallel. (b) For \(\lambda=0\) the solution of \(y^{\prime \prime}=0\) is \(y=c_{1} x+c_{2}\). From the first boundary condition we have $$y(-\pi)=-c_{1} \pi+c_{2}=y(\pi)=c_{1} \pi+c_{2}$$ or \(2 c_{1} \pi=0 .\) Thus, \(c_{1}=0\) and \(y=c_{2} .\) This constant solution is seen to satisfy the boundary-value problem. For \(\lambda=-\alpha^{2}<0\) we have \(y=c_{1} \cosh \alpha x+c_{2} \sinh \alpha x .\) In this case the first boundary condition gives $$\begin{aligned} y(-\pi) &=c_{1} \cosh (-\alpha \pi)+c_{2} \sinh (-\alpha \pi) \\ &=c_{1} \cosh \alpha \pi-c_{2} \sinh \alpha \pi \\ &=y(\pi)=c_{1} \cosh \alpha \pi+c_{2} \sinh \alpha \pi \end{aligned}$$ or \(2 c_{2} \sinh \alpha \pi=0 .\) Thus \(c_{2}=0\) and \(y=c_{1} \cosh \alpha x .\) The second boundary condition implies in a similar fashion that \(c_{1}=0 .\) Thus, for \(\lambda<0\), the only solution of the boundary-value problem is \(y=0\). For \(\lambda=\alpha^{2}>0\) we have \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x .\) The first boundary condition implies $$\begin{aligned} y(-\pi) &=c_{1} \cos (-\alpha \pi)+c_{2} \sin (-\alpha \pi) \\ &=c_{1} \cos \alpha \pi-c_{2} \sin \alpha \pi \\ &=y(\pi)=c_{1} \cos \alpha \pi+c_{2} \sin \alpha \pi \end{aligned}$$ or \(2 c_{2} \sin \alpha \pi=0 .\) Similarly, the second boundary condition implies \(2 c_{1} \alpha \sin \alpha \pi=0 .\) If \(c_{1}=c_{2}=0\) the solution is \(y=0 .\) However, if \(c_{1} \neq 0\) or \(c_{2} \neq 0,\) then \(\sin \alpha \pi=0,\) which implies that \(\alpha\) must be an integer, \(n\) .Therefore, for \(c_{1}\) and \(c_{2}\) not both \(0, y=c_{1} \cos n x+c_{2} \sin n x\) is a nontrivial solution of the boundary-value problem. since \(\cos (-n x)=\cos n x\) and \(\sin (-n x)=-\sin n x,\) we may assume without loss of generality that the eigenvalues are \(\lambda_{n}=\alpha^{2}=n^{2},\) for \(n\) a positive integer. The corresponding eigenfunctions are \(y_{n}=\cos n x\) and \(y_{n}=\sin n x\) c.

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 \\].

From \(x^{\prime \prime}+4 x=-5 \sin 2 t+3 \cos 2 t, x(0)=-1,\) and \(x^{\prime}(0)=1\) we obtain \(x_{c}=c_{1} \cos 2 t+c_{2} \sin 2 t, x_{p}=\) \(\frac{3}{4} t \sin 2 t+\frac{5}{4} t \cos 2 t,\) and \\[x=-\cos 2 t-\frac{1}{8} \sin 2 t+\frac{3}{4} t \sin 2 t+\frac{5}{4} t \cos 2 t\\]

In the case when \(\lambda=-\alpha^{2}<0,\) the solution of the differential equation is \(y=c_{1} \cosh \alpha x+c_{2} \sinh \alpha x .\) The condition \(y(0)=0\) gives \(c_{1}=0\) The condition \(y(1)-\frac{1}{2} y^{\prime}(1)=0 \quad\) applied to \(y=c_{2} \sinh \alpha x\) gives \(c_{2}\left(\sinh \alpha-\frac{1}{2} \alpha \cosh \alpha\right)=0 \quad\) or \(\tanh \alpha=\frac{1}{2} \alpha . \quad\) As can be seen from the figure, the graphs of \(y=\tanh x\) and \(y=\frac{1}{2} x\) intersect at a single point with approximate \(x\) -coordinate \(\alpha_{1}=1.915 .\) Thus, there is a single negative eigenvalue \(\lambda_{1}=-\alpha_{1}^{2} \approx-3.667\) and the corresponding eigenfuntion is \(y_{1}=\sinh 1.915 x\). For \(\lambda=0\) the only solution of the boundary-value problem is \(y=0\) For \(\lambda=\alpha^{2}>0\) the solution of the differential equation is \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x .\) The condition \(y(0)=0\) gives \(c_{1}=0,\) so \(y=c_{2} \sin \alpha x .\) The condition \(y(1)-\frac{1}{2} y^{\prime}(1)=0\) gives \(c_{2}\left(\sin \alpha-\frac{1}{2} \alpha \cos \alpha\right)=0,\) so the eigenvalues are \(\lambda_{n}=\alpha_{n}^{2}\) when \(\alpha_{n}, n=2,3,4, \ldots,\) are the positive roots of \(\tan \alpha=\frac{1}{2} \alpha .\) Using a CAS we find that the first three values of \(\alpha\) are \(\alpha_{2}=4.27487, \alpha_{3}=7.59655,\) and \(\alpha_{4}=10.8127 .\) The first three eigenvalues are then \(\lambda_{2}=\alpha_{2}^{2}=18.2738, \lambda_{3}=\alpha_{3}^{2}=57.7075,\) and \(\lambda_{4}=\alpha_{4}^{2}=116.9139\) with corresponding eigenfunctions \(y_{2}=\sin 4.27487 x, y_{3}=\sin 7.59655 x,\) and \(y_{4}=\sin 10.8127 x\).

If the constant \(-c_{1}^{2}\) is used instead of \(c_{1}^{2},\) then, using partial fractions \\[ y=-\int \frac{d x}{x^{2}-c_{1}^{2}}=-\frac{1}{2 c_{1}} \int\left(\frac{1}{x-c_{1}}-\frac{1}{x+c_{1}}\right) d x=\frac{1}{2 c_{1}} \ln \left|\frac{x+c_{1}}{x-c_{1}}\right|+c_{2} \\] Alternatively, the inverse hyperbolic tangent can be used.
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