Problem 53
Question
Find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$ y \quad c_{1} \cos 8 x+c_{2} \sin 8 x $$
Step-by-Step Solution
Verified Answer
The differential equation is \( y'' + 64y = 0 \).
1Step 1: Identify the Form of the Solution
The general solution given is \( y = c_1 \cos(8x) + c_2 \sin(8x) \). This form corresponds to solutions of a linear differential equation with constant coefficients where the roots of its characteristic equation are complex numbers.
2Step 2: Determine the Roots from the Given Solution
The given solution involves \(\cos(8x)\) and \(\sin(8x)\), which correspond to complex roots of the form \(8i\) and \(-8i\) for the characteristic equation. This implies that the characteristic equation has the form \((r^2 + 8^2) = 0\).
3Step 3: Write the Characteristic Equation
From the complex roots \(r = \pm 8i\), we derive the characteristic equation by setting it as \(r^2 + (8)^2 = 0\), which simplifies to \(r^2 + 64 = 0\).
4Step 4: Construct the Differential Equation
The characteristic equation \(r^2 + 64 = 0\) corresponds to the differential equation: \(y'' + 64y = 0\). This is the homogeneous linear differential equation with constant coefficients that has the given general solution.
Key Concepts
Homogeneous Linear Differential EquationConstant CoefficientsCharacteristic EquationComplex Roots
Homogeneous Linear Differential Equation
A homogeneous linear differential equation is a type of differential equation where the function and all its derivatives appear linearly. The term "homogeneous" indicates that the equation is set to zero instead of being set against a function. Essentially, these equations describe how the solution behaves when it is scaled or shifted.
For example, if our solution is \(y = c_1 \cos(8x) + c_2 \sin(8x)\), it implies the underlying differential equation makes use of linearity and homogeneity.
These types of equations are significant because they model systems where initial conditions dictate the system's response, without external forces acting upon it. These equations are mostly used in fields such as physics and engineering.
For example, if our solution is \(y = c_1 \cos(8x) + c_2 \sin(8x)\), it implies the underlying differential equation makes use of linearity and homogeneity.
These types of equations are significant because they model systems where initial conditions dictate the system's response, without external forces acting upon it. These equations are mostly used in fields such as physics and engineering.
Constant Coefficients
In a differential equation, constant coefficients signify that the coefficients of the derivatives do not change and are constant numbers. Having constant coefficients simplifies both the solving process and analyzing the solutions.
This concept is evident in the equation \(y'' + 64y = 0\). Here, both the second derivative term \(y''\) and the function \(y\) are multiplied by constants (in this case, 1 and 64 respectively).
Equations with constant coefficients are popular because they consistently deliver solutions that can be expressed in terms of exponential functions, sines, and cosines, which are familiar and manageable.
This concept is evident in the equation \(y'' + 64y = 0\). Here, both the second derivative term \(y''\) and the function \(y\) are multiplied by constants (in this case, 1 and 64 respectively).
Equations with constant coefficients are popular because they consistently deliver solutions that can be expressed in terms of exponential functions, sines, and cosines, which are familiar and manageable.
Characteristic Equation
The characteristic equation is a key concept that helps to solve linear differential equations with constant coefficients. This equation is formed by assuming solutions of the differential equation of the form \(e^{rx}\), where \(r\) is a complex number.
For an equation like \(y'' + 64y = 0\), the characteristic equation is found by replacing derivatives with powers of \(r\). In this specific case, we have \(r^2 + 64 = 0\). Solving for \(r\) gives the roots \(\pm 8i\).
The roots of the characteristic equation provide crucial information about the nature of the differential equation's solutions, such as whether they involve exponential growth/decay or oscillatory behavior.
For an equation like \(y'' + 64y = 0\), the characteristic equation is found by replacing derivatives with powers of \(r\). In this specific case, we have \(r^2 + 64 = 0\). Solving for \(r\) gives the roots \(\pm 8i\).
The roots of the characteristic equation provide crucial information about the nature of the differential equation's solutions, such as whether they involve exponential growth/decay or oscillatory behavior.
Complex Roots
When the characteristic equation of a differential equation has complex roots, it indicates that the solutions will involve trigonometric functions.
In our exercise, the characteristic equation has roots \(r = \pm 8i\). Complex roots generally come in conjugate pairs and contribute to solutions expressed with \(\cos\) and \(\sin\) functions as seen in \(y = c_1 \cos(8x) + c_2 \sin(8x)\).
These trigonometric solutions signify oscillatory motion, commonly found in scenarios like alternating currents or mechanical vibrations. Understanding complex roots allows us to anticipate the behavior and form of a solution in systems modeled by differential equations.
In our exercise, the characteristic equation has roots \(r = \pm 8i\). Complex roots generally come in conjugate pairs and contribute to solutions expressed with \(\cos\) and \(\sin\) functions as seen in \(y = c_1 \cos(8x) + c_2 \sin(8x)\).
These trigonometric solutions signify oscillatory motion, commonly found in scenarios like alternating currents or mechanical vibrations. Understanding complex roots allows us to anticipate the behavior and form of a solution in systems modeled by differential equations.
Other exercises in this chapter
Problem 52
Find the steady-state current in an \(L R C\) -series circuit when \(L=\frac{1}{2} \mathrm{~h}, R=20 \Omega, C=0.001 \mathrm{f}\), and \(E(t)=100 \sin 60 t+\) \
View solution Problem 52
In Problems 49-58 find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$ y \quad c_{1} e^{-10 x}+c_{2} x
View solution Problem 53
The initial conditions \(y(0)=y_{0}, y^{\prime}(0)=y_{1}\), apply to each of the following differential equations: $$ \begin{aligned} &x^{2} y^{\prime \prime}=0
View solution Problem 53
Find the charge on the capacitor in an \(L R C\) -series circuit when \(L=\frac{1}{2} \mathrm{~h}, R=10 \Omega, C=0.01 \mathrm{f}, E(t)=150 \mathrm{~V}, q(0)=1
View solution