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Chapter 8: Problem 4

\(y^{\prime \prime}-13 y^{\prime}+40 y=0, y(0)=0, y^{\prime}(0)=-2\)

Short Answer

Expert verified
The solution is \( y(x) = \frac{2}{3} e^{5x} - \frac{2}{3} e^{8x} \)

Step by step solution

01

Form the Characteristic Equation

To solve the differential equation, form the characteristic equation associated with it: \[ r^2 - 13r + 40 = 0 \]
02

Solve the Characteristic Equation

Solve the quadratic equation \( r^2 - 13r + 40 = 0 \). This can be done by factoring: \[ (r - 5)(r - 8) = 0 \]So, the solutions are: \[ r = 5 \]\[ r = 8 \]
03

Write the General Solution

Using the roots obtained, write the general solution of the differential equation: \[ y(x) = C_1 e^{5x} + C_2 e^{8x} \]
04

Apply Initial Conditions

Use the given initial conditions \( y(0) = 0 \) and \( y^{\text{\textquoteleft}}(0) = -2 \) to find the constants. First, apply \( y(0) = 0 \): \[ y(0) = C_1 e^{0} + C_2 e^{0} = C_1 + C_2 = 0 \]So, \[ C_1 = -C_2 \]
05

Differentiate and Apply Second Initial Condition

Differentiate the general solution to use the second initial condition: \[ y^{\text{\textquoteleft}}(x) = 5C_1 e^{5x} + 8C_2 e^{8x} \]Apply \( y^{\text{\textquoteleft}}(0) = -2 \): \[ y^{\text{\textquoteleft}}(0) = 5C_1 + 8C_2 = -2 \]Using \( C_1 = -C_2 \): \[ 5(-C_2) + 8C_2 = -2 \]\[ -5C_2 + 8C_2 = -2 \]\[ 3C_2 = -2 \]\[ C_2 = -\frac{2}{3} \]Therefore, \[ C_1 = \frac{2}{3} \]
06

Write the Final Solution

Substitute the values of \( C_1 \) and \( C_2 \) back into the general solution to get: \[ y(x) = \frac{2}{3} e^{5x} - \frac{2}{3} e^{8x} \]

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Key Concepts

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

Characteristic Equation
A characteristic equation is derived from a second-order linear differential equation with constant coefficients. For the equation given, \( y'' - 13y' + 40y = 0 \), the characteristic equation is found by assuming a solution of the form \( y = e^{rx} \). By substituting \( y = e^{rx} \) into the differential equation, we obtain the characteristic equation:
\[ r^2 - 13r + 40 = 0 \]
This equation is crucial as it helps in finding the roots that determine the general solution of the differential equation. Solving this quadratic equation will yield the values of \( r \) that are used to form the general solution.
Initial Conditions
Initial conditions are given values at which the solution and/or its derivative are specified. They are essential for finding the particular solution from the general solution.
In our exercise, the initial conditions are:
  • \( y(0) = 0 \)
  • \( y'(0) = -2 \)
These initial conditions allow us to solve for the constants \( C_1 \) and \( C_2 \) in the general solution \( y(x) = C_1 e^{5x} + C_2 e^{8x} \). By applying the initial conditions, we can find specific values for these constants ensuring the solution fits the given initial data.
General Solution
The general solution of a second-order differential equation encompasses all possible solutions and includes arbitrary constants. Given the characteristic equation \( r^2 - 13r + 40 = 0 \), solving it through factoring (or the quadratic formula) will give us the roots, \( r = 5 \) and \( r = 8 \).
These roots guide us to the general solution:
\[ y(x) = C_1 e^{5x} + C_2 e^{8x} \]
Here, \( C_1 \) and \( C_2 \) are constants that can be determined using the initial conditions. The exponential terms \( e^{5x} \) and \( e^{8x} \) arise from the values of \( r \). This form is general because it accounts for any possible solution for different initial values.
Factoring Quadratic Equation
To solve the characteristic equation \( r^2 - 13r + 40 = 0 \), we can factor it. Factoring is a method of breaking down the polynomial into simpler terms that when multiplied together yield the original equation.
In this case:
\[ (r - 5)(r - 8) = 0 \]
By setting each factor to zero, we solve for the roots:
  • \( r - 5 = 0 \Rightarrow r = 5 \)
  • \( r - 8 = 0 \Rightarrow r = 8 \)
Factoring the quadratic equation is essential as it gives us the critical values (roots) that we use to form the exponential functions in the general solution of the differential equation. Understanding how to factor quadratic equations is a fundamental skill when working with second-order differential equations.

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Most popular questions from this chapter

\(x^{\prime}-y=0, y^{\prime}+x=f(t), x(0)=y(0)=0\), where \(f(t)=\left\\{\begin{array}{l}\sin t, 0 \leq t<\pi \\ 0, t \geq \pi\end{array}\right.\)

Given that \(X(s)=\mathcal{L}\\{f(t)\\}=\frac{s}{s^{2}+1}+\frac{1}{s^{2}+1}\) \(\times e^{-s \pi / 2}\), show that \(x(t)=\mathcal{L}^{-1}\\{X(s)\\}=\) \(\frac{1}{2} t \sin t+\frac{1}{2}(-\cos t-(t-\pi / 2) \sin t) \mathcal{U}(t-\pi / 2)\).

\(y^{\prime \prime}+9 \pi^{2} y=f(t), f(t)=\delta(t-1)-\delta(t-2)\), \(0 \leq t<3\), and \(f(t)=f(t-3)\) if \(t \geq 3, y(0)=\) \(0, y^{\prime}(0)=1\)

(The Stiffness Matrix) (a) Show that we can write the system in matrix form $$ \begin{aligned} &m_{1} \frac{d^{2} x}{d t^{2}}=-k_{1} x+k_{2}(y-x) \\ &m_{2} \frac{d^{2} y}{d t^{2}}=-k_{2}(y-x) \end{aligned} $$ in matrix form \(\mathbf{M x ^ { \prime \prime }}=\mathbf{K x}\), where \(\mathbf{M}=\) \(\left(\begin{array}{cc}m_{1} & 0 \\ 0 & m_{2}\end{array}\right)\) is the mass matrix and \(\mathbf{K}=\) \(\left(\begin{array}{cc}-\left(k_{1}+k_{2}\right) & k_{2} \\ k_{2} & -k_{2}\end{array}\right)\) is the stiffness matrix. (b) Show that we can write \(\mathbf{M x} \mathbf{x}^{\prime \prime}=\mathbf{K x}\) as \(\mathbf{x}^{\prime \prime}=\mathbf{A x}\), where \(\mathbf{A}=\mathbf{M}^{-1} \mathbf{K}\). (Why does \(\mathbf{M}^{-1}\) exist?) (c) Assume that \(\mathbf{x}(t)=\mathbf{v} e^{\alpha t}\) is a solution of \(\mathbf{x}^{\prime \prime}=\mathbf{A x}\). Show that \(\mathbf{x}(t)=\mathbf{v} e^{\alpha t}\) satisfies the system if and only if \(\alpha^{2}=\lambda\), where \(\lambda\) is an eigenvalue of \(\mathbf{A}\) and \(\mathbf{v}\) is an associated eigenvector. (d) The eigenvalues of \(\mathbf{A}\) are typically negative real numbers. Show that two solutions of the system are \(\mathbf{x}_{1}=\mathbf{v} \cos \omega t\) and \(\mathbf{x}_{2}=\mathbf{v} \sin \omega t\), where \(\alpha^{2}=\lambda=-\omega^{2}\) or \(\alpha=\pm \omega i\).

\(f(t)=\delta(t-2)-2 \delta(t-1), 0 \leq t<3, f(t)=\) \(f(t-3)\) if \(t \geq 3\)
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