Problem 35
Question
In Problems 33-38, find a homogeneous Cauchy-Euler differential equation whose general solution is given. $$ y=c_{1} x^{-3}+c_{2} x^{-3} \ln x $$
Step-by-Step Solution
Verified Answer
The differential equation is \(x^2 y'' + 6x y' + 9y = 0\).
1Step 1: Identify the form of the solution
The given solution is \( y = c_1 x^{-3} + c_2 x^{-3} \ln x \). We recognize that this is a linear combination of functions, one of which includes a logarithmic term, indicating a repeated root.
2Step 2: Recognize the solution structure
The solution \( c_1 x^{-3} + c_2 x^{-3} \ln x \) suggests that it arises from a Cauchy-Euler differential equation where \( m_1 = m_2 = -3 \). The presence of \( \ln x \) suggests a repeated real root.
3Step 3: Write the characteristic equation
For a Cauchy-Euler equation, the characteristic equation related to the solution structure is usually found by assuming \( y = x^m \), which leads to an equation \( m(m-1) + am + b = 0 \). Given the repeated root \( m = -3 \), the characteristic equation must be \((m + 3)^2 = 0\).
4Step 4: Expand the characteristic equation
Expanding \((m + 3)^2 = 0\) gives \(m^2 + 6m + 9 = 0\). This corresponds to the differential equation coefficients.
5Step 5: Form the differential equation
The homogeneous Cauchy-Euler equation derived from \(m^2 + 6m + 9 = 0\) will be:\[x^2 y'' + 6x y' + 9y = 0\].
Key Concepts
Homogeneous Differential EquationCharacteristic EquationRepeated Roots in Differential Equations
Homogeneous Differential Equation
A homogeneous differential equation is a type of differential equation where every term is a function of the dependent variable and its derivatives. This means such equations can be equated to zero. In mathematics, solving these equations often involves finding a general solution that satisfies the entire equation. By equating the equation to zero, the dependent variable's influence can be isolated and analyzed more systematically.
Cauchy-Euler differential equations fall within the category of homogeneous differential equations. These differential equations have a unique format which is linked to their coefficients being functions of the independent variable in a specific form. This specific form arises due to the power-type relationship defined in the differential equation.
Cauchy-Euler differential equations fall within the category of homogeneous differential equations. These differential equations have a unique format which is linked to their coefficients being functions of the independent variable in a specific form. This specific form arises due to the power-type relationship defined in the differential equation.
- This format helps us conveniently address solution behavior, particularly when the roots of the characteristic equation are real or complex numbers.
- The Cauchy-Euler equations are especially useful in solving problems where these variable coefficients mimic the setup of practical engineering problems.
Characteristic Equation
The characteristic equation is key in solving differential equations, especially those of Cauchy-Euler type. To identify this equation, we assume a solution form and derive the equation by substituting it into the differential equation. The process often involves the substitution of expressions of the dependent variable in terms of powers across their derivatives.
For the Cauchy-Euler differential equation, this assumption leads to an equation based on an invariant power followed by a function or polynomial of the root of the characteristic equation.
For the Cauchy-Euler differential equation, this assumption leads to an equation based on an invariant power followed by a function or polynomial of the root of the characteristic equation.
- The characteristic roots indicate the form of the general solution, whether it includes pure exponential terms, oscillations, or combinations thereof.
- The identification of the characteristic equation emerges from known assumptions, typically following the pattern of forming a quadratic or higher degree polynomial from these assumed expressions.
Repeated Roots in Differential Equations
Repeated roots occur when the characteristic equation has multiple roots of the same value. This situation presents unique challenges because the usual exponential solutions need augmentation to accommodate these repetitions. When encountering repeated roots, mathematicians turn to modified solutions to ensure that they span the solution space properly.
In the specific case of Cauchy-Euler equations, repeated roots involving logarithmic terms appear. These logarithmic components ensure completeness in the solution representation. A common method of expressing the solution involves adding terms that differ by multiplicative constants and possessing logarithmic relationships along with the initial repeated form:
In the specific case of Cauchy-Euler equations, repeated roots involving logarithmic terms appear. These logarithmic components ensure completeness in the solution representation. A common method of expressing the solution involves adding terms that differ by multiplicative constants and possessing logarithmic relationships along with the initial repeated form:
- For each repeated root, additional terms usually involve multiplications with polynomials which increase the degree of the expression.
- In real-world applications, these solutions cater to multiple physical scenarios by addressing different dynamic system responses.
Other exercises in this chapter
Problem 35
(a) Verify that \(y_{P_{1}}=3 e^{2 x}\) and \(y_{p_{1}}=x^{2}+3 x\) are, respectively, particular solutions of $$ y^{\prime \prime}-6 y^{\prime}+5 y=-9 e^{2 x}
View solution Problem 35
In Problems 33-38, solve the given differential equation subject to the indicated conditions. $$ y^{\prime \prime}-y=x+\sin x, y(0)=2, y^{\prime}(0)=3 $$
View solution Problem 35
$$ \text { In Problems 27-36, solve the given initial-value problem. } $$ $$ \begin{aligned} &y^{\prime \prime \prime}-2 y^{\prime \prime}+y^{\prime}=2-24 e^{x}
View solution Problem 35
(a) Verify that \(y_{p_{1}}=3 e^{2 x}\) and \(y_{p_{2}}=x^{2}+3 x\) are, respectively, particular solutions of $$ y^{\prime \prime}-6 y^{\prime}+5 y=-9 e^{2 x}
View solution