Problem 52
Question
In Problems 49-58 find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$ y \quad c_{1} e^{-10 x}+c_{2} x e^{-10 x} $$
Step-by-Step Solution
Verified Answer
The differential equation is \( y'' + 20y' + 100y = 0 \).
1Step 1: Identifying the Form of the General Solution
The given general solution is \( y = c_1 e^{-10x} + c_2 x e^{-10x} \). Notice that it consists of terms involving \( e^{-10x} \) and \( x e^{-10x} \). This suggests the characteristic equation of the differential equation has a repeated root.
2Step 2: Determine the Repeated Root
Since the terms \( e^{-10x} \) and \( x e^{-10x} \) are present, the characteristic equation must have a root of \( r = -10 \) with multiplicity 2. This means the differential equation has a characteristic equation of the form \((r + 10)^2 = 0\).
3Step 3: Construct the Characteristic Equation
The characteristic equation derived from the general solution is \((r + 10)^2 = 0\), which simplifies to \(r^2 + 20r + 100 = 0\).
4Step 4: Write the Differential Equation
The characteristic equation \(r^2 + 20r + 100 = 0\) corresponds to a homogeneous linear differential equation with constant coefficients given by: \[ y'' + 20y' + 100y = 0 \]
Key Concepts
Constant CoefficientsCharacteristic EquationRepeated Roots
Constant Coefficients
When dealing with homogeneous linear differential equations, one key feature is the constant coefficients. This means the coefficients of all derivatives in the equation are constants, not variables. Let's take an example:
- Consider the differential equation: \(y'' + 3y' + 2y = 0\). Here, the coefficients 3 and 2 are constants.
Having constant coefficients simplifies the process of finding general solutions. It allows us to use methods like the characteristic equation to solve these equations more systematically. Such equations typically arise in physical phenomena like mechanics and electrical circuits, where proportional relationships between derivatives are governed by fixed factors.
In the exercise, the differential equation has constant coefficients derived from the characteristic polynomial, ensuring the form \(y'' + 20y' + 100y = 0\) matches the general solution.
- Consider the differential equation: \(y'' + 3y' + 2y = 0\). Here, the coefficients 3 and 2 are constants.
Having constant coefficients simplifies the process of finding general solutions. It allows us to use methods like the characteristic equation to solve these equations more systematically. Such equations typically arise in physical phenomena like mechanics and electrical circuits, where proportional relationships between derivatives are governed by fixed factors.
In the exercise, the differential equation has constant coefficients derived from the characteristic polynomial, ensuring the form \(y'' + 20y' + 100y = 0\) matches the general solution.
Characteristic Equation
The characteristic equation is crucial in solving linear differential equations with constant coefficients. It transforms a differential equation into an algebraic one, making it easier to solve. Here's the process:
- For a second-order differential equation like \(y'' + 20y' + 100y = 0\), assume a solution of the form \(y = e^{rx}\).
- Substitute \(y = e^{rx}\) into the differential equation.
- This substitution leads to an algebraic equation in \(r\), known as the characteristic equation.
In our exercise, we derived \((r + 10)^2 = 0\) as the characteristic equation. Solving this, we find \(r = -10\) with multiplicity 2. The roots of this characteristic equation reveal the behavior of solutions to the corresponding differential equation.
- For a second-order differential equation like \(y'' + 20y' + 100y = 0\), assume a solution of the form \(y = e^{rx}\).
- Substitute \(y = e^{rx}\) into the differential equation.
- This substitution leads to an algebraic equation in \(r\), known as the characteristic equation.
In our exercise, we derived \((r + 10)^2 = 0\) as the characteristic equation. Solving this, we find \(r = -10\) with multiplicity 2. The roots of this characteristic equation reveal the behavior of solutions to the corresponding differential equation.
Repeated Roots
Repeated roots occur when the characteristic equation has roots with multiplicity greater than one. This means the solution requires extra terms for each repeat of the root to account for all conditions. In our situation, the characteristic equation \((r + 10)^2 = 0\) has \(r = -10\) as a repeated root with multiplicity 2.
For each repeated root, the associated general solution includes additional terms involving powers of \(x\). For a repeated root \(r\) with multiplicity, the terms are:
For each repeated root, the associated general solution includes additional terms involving powers of \(x\). For a repeated root \(r\) with multiplicity, the terms are:
- \(c_1 e^{rx}\)
- \(c_2 x e^{rx}\)
Other exercises in this chapter
Problem 52
Find a Cauchy-Euler differential equation of lowest order with real coefficients if it is known that 2 and \(1-i\) are two roots of its auxiliary equation.
View solution 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 53
Find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$ y \quad c_{1} \cos 8 x+c_{2} \sin 8 x $$
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