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Chapter 3: Problem 1

We have \(y_{1}^{\prime}=y_{1}^{\prime \prime}=e^{x},\) so \\[ \left(y_{1}^{\prime \prime}\right)^{2}=\left(e^{x}\right)^{2}=e^{2 x}=y_{1}^{2} \\] Also, \(y_{2}^{\prime}=-\sin x\) and \(y_{2}^{\prime \prime}=-\cos x,\) so \\[ \left(y_{2}^{\prime \prime}\right)^{2}=(-\cos x)^{2}=\cos ^{2} x=y_{2}^{2} \\] However, if \(y=c_{1} y_{1}+c_{2} y_{2},\) we have \(\left(y^{\prime \prime}\right)^{2}=\left(c_{1} e^{x}-c_{2} \cos x\right)^{2}\) and \(y^{2}=\left(c_{1} e^{x}+c_{2} \cos x\right)^{2} .\) Thus \(\left(y^{\prime \prime}\right)^{2} \neq y^{2}.\)

Short Answer

Expert verified
\( y^2 \) does not equal \( (y'')^2 \) for the linear combination.

Step by step solution

01

Understand the Given Functions

We are given two systems of equations:1. \( y_1'' = e^x \) and \( y_1' = e^x \), indicating that \( y_1'' = y_1' \).2. \( y_2' = -\sin x \) and \( y_2'' = -\cos x \).We need to verify and interpret expressions for \( y_1^2 \) and \( y_2^2 \).
02

Square the Second Derivatives

For \( y_1 \), \[ (y_1'')^2 = (e^x)^2 = e^{2x} = y_1^2. \]For \( y_2 \), \[ (y_2'')^2 = (-\cos x)^2 = \cos^2 x = y_2^2. \]
03

Analyze the Linear Combination

Consider \( y = c_1 y_1 + c_2 y_2 \). Take the second derivative:\[ y'' = c_1 y_1'' + c_2 y_2'' = c_1 e^x - c_2 \cos x. \]

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Key Concepts

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

Second Derivatives
In differential equations, the second derivative gives us valuable information about the curvature and concavity of a function. For instance, in our exercise, we are given two different scenarios for second derivatives:
  • For the function with the second derivative equal to the exponential function, we have \( y_1'' = e^x \). This tells us the rate at which the slope itself is changing over time, growing exponentially.
  • When examining trigonometric functions, like \( y_2'' = -\cos x \), it tells us how the oscillation or wave of the function is bending, as cosine oscillates between -1 and 1.
Exploring these derivatives helps us understand how these two types of functions behave in terms of growth and oscillation, providing insights into their shapes and behaviors.
This is why squaring the second derivative, as done in the problem, leads to reflections on their inherent characteristics, such as symmetry in trigonometric functions and ever-increasing growth in exponential ones.
Linear Combination
A linear combination refers to the process of combining different functions or variables with constant multipliers to create a new function. In our problem, we have:
The linear combination is given by \( y = c_1 y_1 + c_2 y_2 \). Here, \( c_1 \) and \( c_2 \) are constants, which allows us to mix two functions to form a new one. Understanding linear combinations is crucial in differential equations as they often model real-world phenomena by mixing different components.
  • The derivative of this combination, \( y'' = c_1 y_1'' + c_2 y_2'' \), shows us how the new function's curvature will be derived from its components, influencing the overall system's behavior.
  • This also means that unlike the constituent functions, the square of the second derivative, \( (y'')^2 \), is not necessarily equivalent to the square of the combined function, \( y^2 \), revealing insights into how these mixed terms can interact non-linearly.
Exponential Functions
Exponential functions are a key concept in differential equations due to their unique properties of growth and decay. The function \( e^x \), which frequently appears in calculus and differential equations, is significant for several reasons:
  • It models constant proportional growth, meaning the rate of change of the function is consistent with its value at any point.
  • When differentiated, it remains unchanged, i.e., the derivative of \( e^x \) is \( e^x \), highlighting a key feature of exponential functions that makes them predictable and valuable in modeling scenarios like population growth and radioactive decay.
Their importance in differential equations lies in how they simplify solving linear differential equations, often serving as solutions that exploit their growth characteristics. As seen in the exercise, \( y_1'' = e^x \), this function's behavior can describe numerous physical and theoretical systems effectively.

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Most popular questions from this chapter

If \(\lambda=\alpha^{2}=P / E I,\) then the solution of the differential equation is \\[ y=c_{1} \cos \alpha x+c_{2} \sin \alpha x+c_{3} x+c_{4} \\]. The conditions \(y(0)=0, y^{\prime \prime}(0)=0\) yield, in turn, \(c_{1}+c_{4}=0\) and \(c_{1}=0 .\) With \(c_{1}=0\) and \(c_{4}=0\) the solution is \(y=c_{2} \sin \alpha x+c_{3} x .\) The conditions \(y(L)=0, y^{\prime \prime}(L)=0,\) then yield $$c_{2} \sin \alpha L+c_{3} L=0 \quad \text { and } \quad c_{2} \sin \alpha L=0.$$Hence, nontrivial solutions of the problem exist only if \(\sin \alpha L=0 .\) From this point on, the analysis is the same as in Example 3 in the text.

The general solution of the differential equation \(y^{\prime \prime}+3 y=6 x\) is \(y=c_{1} \cos \sqrt{3} x+c_{2} \sin \sqrt{3} x+2 x .\) The condition \(y(0)=0\) implies \(c_{1}=0\) and so \(y=c_{2} \sin \sqrt{3} x+2 x .\) The condition \(y(1)+y^{\prime}(1)=0\) implies \(c_{2} \sin \sqrt{3}+2+c_{2} \sqrt{3} \cos \sqrt{3}+2=0\) so \(c_{2}=-4 /(\sin \sqrt{3}+\sqrt{3} \cos \sqrt{3}) .\) The solution is $$y=\frac{-4 \sin \sqrt{3} x}{\sin \sqrt{3}+\sqrt{3} \cos \sqrt{3}}+2 x$$

Solving \(\frac{1}{4} q^{\prime \prime}+20 q^{\prime}+300 q=0\) we obtain \(q(t)=c_{1} e^{-20 t}+c_{2} e^{-60 t} .\) The initial conditions \(q(0)=4\) and \(q^{\prime}(0)=0\) imply \(c_{1}=6\) and \(c_{2}=-2 .\) Thus \\[q(t)=6 e^{-20 t}-2 e^{-60 t}\\] Setting \(q=0\) we find \(e^{40 t}=1 / 3\) which implies \(t<0 .\) Therefore the charge is not 0 for \(t \geq 0\).

(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 \(\lambda \leq 0\) the only solution of the boundary-value problem is \(y=0 .\) For \(\lambda=\alpha^{2}>0\) we have $$y=c_{1} \cos \alpha x+c_{2} \sin \alpha x$$ Now \(y(-\pi)=y(\pi)=0\) implies \\[ \begin{array}{l} c_{1} \cos \alpha \pi-c_{2} \sin \alpha \pi=0 \\ c_{1} \cos \alpha \pi+c_{2} \sin \alpha \pi=0 \end{array} \\] This homogeneous system will have a nontrivial solution when \\[ \left|\begin{array}{rr} \cos \alpha \pi & -\sin \alpha \pi \\ \cos \alpha \pi & \sin \alpha \pi \end{array}\right|=2 \sin \alpha \pi \cos \alpha \pi=\sin 2 \alpha \pi=0 \\] Then \\[ 2 \alpha \pi=n \pi \quad \text { or } \quad \lambda=\alpha^{2}=\frac{n^{2}}{4} ; \quad n=1,2,3, \ldots \\] When \(n=2 k-1\) is odd, the eigenvalues are \((2 k-1)^{2} / 4 .\) since \(\cos (2 k-1) \pi / 2=0\) and \(\sin (2 k-1) \pi / 2 \neq 0\) we see from either equation in (1) that \(c_{2}=0 .\) Thus, the eigenfunctions corresponding to the eigenvalues \((2 k-1)^{2} / 4\) are \(y=\cos (2 k-1) x / 2\) for \(k=1,2,3, \ldots .\) Similarly, when \(n=2 k\) is even, the eigenvalues are \(k^{2}\) with corresponding eigenfunctions \(y=\sin k x\) for \(k=1,2,3, \ldots\)
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