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Magazine Chapter 3: Problem 32
In Problems 31 and 32, solve the given boundary-value problem. $$ x^{2} y^{\prime \prime}-3 x y^{\prime}+5 y=0, y(1)=0, y(e)=1 $$
Short Answer
Expert verified
The particular solution is \( y = \frac{x^2 \sin(\ln x)}{e^2 \sin(1)} \).
Step by step solution
01
Identify the Type of Differential Equation
The differential equation given is a second-order linear homogeneous equation with variable coefficients: \[ x^2 y'' - 3xy' + 5y = 0 \] This form suggests that it could be an Euler-Cauchy equation (or a Cauchy-Euler equation).
02
Apply the Cauchy-Euler Equation Method
For an Euler-Cauchy equation of the form \[ x^2 y'' + axy' + by = 0 \], the typical solution is \[ y = x^m \]. Substituting \( y = x^m \) and its derivatives into the equation will yield an indicial equation.
03
Find the Indicial Equation
Substitute \( y = x^m \), \( y' = mx^{m-1} \), and \( y'' = m(m-1)x^{m-2} \) into the differential equation: \[ x^2(m(m-1)x^{m-2}) - 3x(mx^{m-1}) + 5x^m = 0 \]This simplifies to:\[ m(m-1)x^m - 3mx^m + 5x^m = 0 \] Factor out \( x^m \):\[ (m(m-1) - 3m + 5)x^m = 0 \]The indicial equation is:\[ m^2 - 4m + 5 = 0 \].
04
Solve the Indicial Equation
The indicial equation \( m^2 - 4m + 5 = 0 \) is a quadratic equation. Use the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1, b = -4, c = 5 \).\[ m = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} \]\[ m = \frac{4 \pm 2i}{2} = 2 \pm i \]The roots are complex: \( 2 + i \) and \( 2 - i \).
05
Write the General Solution
For complex roots \( m = \alpha \pm \beta i \), the solution is:\[ y = C_1 x^eta \cos(\beta \ln x) + C_2 x^eta \sin(\beta \ln x) \]Substitute \( \alpha = 2 \) and \( \beta = 1 \):\[ y = x^2 (C_1 \cos(\ln x) + C_2 \sin(\ln x)) \]
06
Apply the Boundary Conditions
Given \( y(1) = 0 \) and \( y(e) = 1 \), substitute into the general solution to find \( C_1 \) and \( C_2 \):1. At \( x = 1 \): \[ 0 = 1^2 (C_1 \cos(\ln 1) + C_2 \sin(\ln 1)) = C_1 \Rightarrow \ C_1 = 0 \]2. At \( x = e \): \[ 1 = e^2 (0 + C_2 \sin(1)) \Rightarrow C_2 = \frac{1}{e^2 \sin(1)} \]
07
Write the Particular Solution
Substitute back into the general solution, using \( C_1 = 0 \) and \( C_2 = \frac{1}{e^2 \sin(1)} \):\[ y = x^2 \left(0 \cdot \cos(\ln x) + \frac{1}{e^2 \sin(1)} \sin(\ln x)\right) = \frac{x^2 \sin(\ln x)}{e^2 \sin(1)} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Euler-Cauchy equation
The Euler-Cauchy equation is a special form of a differential equation known for its property of having variable coefficients. It is typically expressed in the form:
The distinctive feature of this equation is that the coefficients correspond to powers of the independent variable \( x \). This structure makes it suitable for a substitution method, often used to simplify and solve the equation.
In solving an Euler-Cauchy equation, a typical strategy is to propose a solution of the form \( y = x^m \), leading to an indicial equation upon substitution. This equation allows us to identify the possible values or "roots" for \( m \), determining the form of the general solution.
The simplicity and elegance of the Euler-Cauchy equation make it quite appealing in exploring boundary-value problems, where a solution is expected to meet certain specific conditions at the boundaries of an interval.
- \( x^2 y'' + axy' + by = 0 \)
The distinctive feature of this equation is that the coefficients correspond to powers of the independent variable \( x \). This structure makes it suitable for a substitution method, often used to simplify and solve the equation.
In solving an Euler-Cauchy equation, a typical strategy is to propose a solution of the form \( y = x^m \), leading to an indicial equation upon substitution. This equation allows us to identify the possible values or "roots" for \( m \), determining the form of the general solution.
The simplicity and elegance of the Euler-Cauchy equation make it quite appealing in exploring boundary-value problems, where a solution is expected to meet certain specific conditions at the boundaries of an interval.
Second-order differential equation
Second-order differential equations involve second derivatives and are common in modeling physical systems and phenomena such as motion and oscillations. They are generally expressed in the form:
When these equations have constant coefficients, they are easier to solve, often using characteristic equations. However, in the case of the Euler-Cauchy equation, the coefficients are variable, making the resolution process more complex.
Second-order equations can be homogeneous or non-homogeneous. In the homogeneous case, like in our problem, the right-hand side equals zero \( g(x) = 0 \). This property implies that solutions are dependent on initial or boundary conditions provided.
In our example, the differential equation was:
- \( y'' + p(x)y' + q(x)y = g(x) \)
When these equations have constant coefficients, they are easier to solve, often using characteristic equations. However, in the case of the Euler-Cauchy equation, the coefficients are variable, making the resolution process more complex.
Second-order equations can be homogeneous or non-homogeneous. In the homogeneous case, like in our problem, the right-hand side equals zero \( g(x) = 0 \). This property implies that solutions are dependent on initial or boundary conditions provided.
In our example, the differential equation was:
- \( x^2 y'' - 3x y' + 5y = 0 \)
Complex roots in differential equations
Complex roots in the solutions of differential equations often arise when the characteristic equation has terms leading to the discriminant being negative. Such situations require a special form of the solution to cater to the complex component.
For instance, consider the indicial equation from our Euler-Cauchy problem:
When dealing with complex roots, \( \alpha \pm \beta i \), the general solution structure includes trigonometric functions, and a typical expression would be:
This form accommodates the oscillatory nature introduced by the imaginary part \( \beta \), resulting in cosine and sine components dependent on \( \ln x \). This way, even though the actual roots aren't real, the solution is transformed back into real-valued functions, which can satisfy the given boundary conditions in problems.
For instance, consider the indicial equation from our Euler-Cauchy problem:
- \( m^2 - 4m + 5 = 0 \)
When dealing with complex roots, \( \alpha \pm \beta i \), the general solution structure includes trigonometric functions, and a typical expression would be:
- \( y = x^\alpha (C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x)) \)
This form accommodates the oscillatory nature introduced by the imaginary part \( \beta \), resulting in cosine and sine components dependent on \( \ln x \). This way, even though the actual roots aren't real, the solution is transformed back into real-valued functions, which can satisfy the given boundary conditions in problems.
