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Magazine Chapter 17: Problem 25
Solve the given initial value problem. \(x^{2} y^{\prime \prime}+3 x y^{\prime}-3 y=0, \quad y(1)=1, y^{\prime}(1)=-1\)
Short Answer
Expert verified
The solution is \( y(x) = 0.5x + 0.5x^{-3} \).
Step by step solution
01
Recognize the Differential Equation Type
The given equation is a second-order linear homogeneous differential equation with variable coefficients. We need to determine whether it falls into any specific category like Cauchy-Euler equations that could simplify the solving process.
02
Change Variables to Simplify
The equation appears to be a Cauchy-Euler equation since it has the form \[ x^2 y'' + 3x y' - 3y = 0. \]By substituting \( x = e^t \) (implying \( dx/dt = e^t \)), we change the variable to transform the differential equation into a constant coefficient form.
03
Transform the Equation
By substituting, we have:\[ y' = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{e^t} \frac{dy}{dt}, \quad y'' = \frac{d}{dt} \left(\frac{1}{e^t} \frac{dy}{dt}\right). \]After simplifying, this leads to \[ u'' + 2u' - 3u = 0, \]where \( u(t) = y(e^t). \) This is now a second-order linear differential equation with constant coefficients.
04
Solution of the Characteristic Equation
The characteristic equation for \( u'' + 2u' - 3u = 0 \) is obtained by assuming a solution of the form \( u(t) = e^{mt} \), yielding the characteristic equation:\[ m^2 + 2m - 3 = 0. \]
05
Solve the Characteristic Equation
Solve the quadratic equation \[ m^2 + 2m - 3 = 0 \] Using the quadratic formula, \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 2, c = -3 \), we find the roots:\[ m_1 = 1, \quad m_2 = -3. \]
06
General Solution of the Transformed Equation
Since the characteristic equation has distinct real roots, the general solution for \( u(t) \) is:\[ u(t) = C_1 e^{t} + C_2 e^{-3t}. \]
07
Transform Back to the Original Variable
Substituting back \( t = \ln(x) \), the general solution in terms of \( x \) becomes:\[ y(x) = C_1 x + C_2 x^{-3}. \]
08
Apply Initial Conditions
Use the initial conditions \( y(1) = 1 \) and \( y'(1) = -1 \).Substitute \( x = 1 \) into the solution:\[ y(1) = C_1 + C_2 = 1. \]For the derivative, \( y' = C_1 - 3C_2 x^{-4} \). Substitute and solve:\[ y'(1) = C_1 - 3C_2 = -1. \]
09
Solve the System of Equations
Solving the system:1. \( C_1 + C_2 = 1 \)2. \( C_1 - 3C_2 = -1 \)Subtracting equation 1 from equation 2 solves to \( 4C_2 = 2 \) implying \( C_2 = 0.5 \).Substitute \( C_2 = 0.5 \) back into equation 1, \( C_1 + 0.5 = 1 \) implies \( C_1 = 0.5 \).
10
Final Specific Solution
With the solved constants, the solution to the initial value problem is:\[ y(x) = 0.5x + 0.5x^{-3}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-Order Linear Differential Equations
Second-order linear differential equations form the backbone of many mathematical models, especially in engineering and physics. These equations involve derivatives up to the second order. A general form looks like:
\[ a(x)y'' + b(x)y' + c(x)y = 0 \]
where \( a(x), b(x), \) and \( c(x) \) are functions of the independent variable \( x \). The goal is to find the function \( y(x) \) that satisfies this equation.
\[ a(x)y'' + b(x)y' + c(x)y = 0 \]
where \( a(x), b(x), \) and \( c(x) \) are functions of the independent variable \( x \). The goal is to find the function \( y(x) \) that satisfies this equation.
- Order and Linearity: The 'order' refers to the highest derivative in the equation, while 'linearity' implies that each term is either a constant or a simple product of a constant with a dependent variable or its derivatives.
- Homogeneous Form: This signifies that the equation equals zero, ensuring that only the terms with the dependent variable and its derivatives appear.
Characteristic Equation
The characteristic equation is a crucial tool in solving linear differential equations, particularly those with constant coefficients. To derive it, we assume a solution of the form \( y = e^{rx} \). By substituting this into the differential equation, a polynomial (called the characteristic polynomial) is formed:
\[ ar^2 + br + c = 0 \]
Here, \( r \) represents a potential root, revealing information about the solution's nature.
\[ ar^2 + br + c = 0 \]
Here, \( r \) represents a potential root, revealing information about the solution's nature.
- Roots Interpretation: Real and distinct roots indicate a general solution composed of exponential terms. Complex roots introduce sine and cosine functions, illustrating oscillatory behavior.
- Finding Solutions: Once the roots are known, they guide the formation of the general solution. In our specific example, the roots \( m_1 = 1 \) and \( m_2 = -3 \) lead to solutions \( u(t) = C_1 e^t + C_2 e^{-3t} \).
Initial Value Problem
In many practical situations, specific conditions are given at an initial point, making the problem an initial value problem. This means finding a solution to a differential equation that satisfies conditions at a particular point in time or space.
- Importance of Initial Conditions: They ensure the solution is 'anchored' at a specific point, allowing for a unique solution among potentially infinite possibilities.
- Application in Cauchy-Euler Equations: In the equation \( x^2y'' + 3xy' - 3y = 0 \), initial conditions \( y(1) = 1 \) and \( y'(1) = -1 \) pin the exact curve needed.
- Solving the System: Using these conditions with the general solution helps solve for integration constants uniquely, giving the specific solution \( y(x) = 0.5x + 0.5x^{-3} \).
Variable Substitution in Differential Equations
Variable substitution is a powerful technique used to simplify complex differential equations by transforming variables into forms that are more manageable.
In the Cauchy-Euler equation \( x^2y'' + 3xy' - 3y = 0 \), substitution \( x = e^t \) is used. This transforms the variable into a similar equation with constant coefficients, radically simplifying the solution process.
In the Cauchy-Euler equation \( x^2y'' + 3xy' - 3y = 0 \), substitution \( x = e^t \) is used. This transforms the variable into a similar equation with constant coefficients, radically simplifying the solution process.
- Why Substitute?: Equations that may not be solvable in their original form might become linear and more approachable after substitution.
- Transformation Process: Substituting \( x = e^t \) yields \( y(t) = y(e^t), y'(x) \) and \( y''(x) \) in terms of \( t \). This turns the equation into a constant coefficient differential equation \( u'' + 2u' - 3u = 0 \).
