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Magazine Chapter 3: Problem 52
Make the substitution \(v=\ln x\) of Problem 51 to find general solutions (for \(x>0\) ) of the Euler equations in Problems. $$ x^{2} y^{\prime \prime}+x y^{\prime}+9 y=0 $$
Short Answer
Expert verified
The general solution is \\(y = x^{-1/2}(C_1 \cos(3 \ln x) + C_2 \sin(3 \ln x))\\).
Step by step solution
01
Make the Substitution
First, we perform the substitution given in the problem: let \(v = \ln x\). This implies \(x = e^v\) and by the chain rule, this gives us \(\frac{dx}{dv} = e^v = x\). The derivatives \(y'\) and \(y''\) in terms of \(v\) are transformed as follows: \(\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}\), and \(\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dv} \cdot \frac{dv}{dx})\).
02
Express the Derivatives
Now let's express the derivatives with respect to \(v\). For the first derivative, calculate \(y' = \frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{1}{x}\). For the second derivative, use: \(y'' = \frac{d}{dx}(\frac{dy}{dv} \cdot \frac{1}{x})\). This will simplify to: \(y'' = \frac{1}{x} \cdot \frac{d}{dv}(\frac{dy}{dv}) - \frac{1}{x^2} \cdot \frac{dy}{dv}\).
03
Substitute into the Equation
Substitute these expressions into the original differential equation: \[ x^2 \left(\frac{1}{x} \cdot \frac{d}{dv}(\frac{dy}{dv}) - \frac{1}{x^2} \cdot \frac{dy}{dv} \right) + x \left( \frac{dy}{dv} \cdot \frac{1}{x} \right) + 9y = 0 \]Simplify using \(x^2 = e^{2v}\) to obtain a simpler equation in terms of \(v\): \[ e^{2v} \frac{d^2y}{dv^2} - e^{v} \frac{dy}{dv} + 9y = 0 \].
04
Solve the Simplified Equation
Notice that because of the term \(e^{2v}\), we can transform this into a constant coefficient equation by assuming a solution of the form \(y = e^{\lambda v}\). Substitute \(y = e^{\lambda v}\) into the equation, which gives us a characteristic equation. This results in a simplification process after which you find the roots of a quadratic characteristic equation.
05
Find the Roots and General Solution
The characteristic equation derived from the previous substitution typically looks like: \ \( \lambda^2 + \text{lower terms} = 0 \, \) solving this will give two roots, say \( \lambda_1 \) and \( \lambda_2 \). The general solution for \(y\) in terms of \(v\) then is \( y = C_1 e^{\lambda_1 v} + C_2 e^{\lambda_2 v} \). Convert back from \(v\) to \(x\) using \(v = \ln x\), resulting in \(y = C_1 x^{\lambda_1} + C_2 x^{\lambda_2}\).
06
Final General Solution
After solving \( \lambda^2 - 1 \lambda + 9 = 0 \), you obtain complex roots: \( \lambda_1 = \alpha + \beta i\) and \( \lambda_2 = \alpha - \beta i \). Translate these in terms of \(x\) to get the general structure: \[y = x^{\alpha}(C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x))\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique in solving differential equations, particularly when dealing with Euler equations. The main idea is to transform a differential equation into a simpler form that is easier to solve. In the presented exercise, we apply the substitution \( v = \ln x \). This changes the problem from a differential equation in terms of \( x \) to one in terms of \( v \). Since \( x = e^v \), this substitution simplifies handling derivatives. Here鈥檚 how it helps:
- Reduces complexity: By switching variables, the equation might become linear or have constant coefficients.
- Simplifies differentiation: It turns complicated expressions into more manageable forms.
Characteristic Equation
When solving differential equations, especially those with constant coefficients, finding the characteristic equation is a crucial step. It emerges from assuming a solution of particular exponential form, typically \( y = e^{\lambda v} \). By substituting this into our differential equation, terms align neatly to reveal a polynomial equation in \( \lambda \). This polynomial is referred to as the characteristic equation. Solving this release the roots, which direct the nature of the general solution:
- Real and distinct roots: Lead directly to exponential solutions.
- Repeated roots: Demand modification with additional polynomial terms.
- Complex roots: Result in oscillatory solutions using sine and cosine functions.
Complex Roots
Complex roots arise when solving the characteristic equation. This situation often happens in Euler equations, resulting in complex solutions that require a different formulation. For instance, with roots of the form \( \lambda_1 = \alpha + \beta i \) and \( \lambda_2 = \alpha - \beta i \), we transform the exponential solutions into trigonometric forms involving sine and cosine. This leads to a general solution for equations like the Euler equation:
- \( y = x^\alpha (C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x)) \)
- This form exploits Euler鈥檚 formula linking complex exponentials to trigonometric functions.
- These solutions represent oscillatory behavior, common in contexts like mechanical vibrations.
General Solution
A general solution is a complete set of possible solutions to a differential equation that involves arbitrary constants. It implicitly includes all specific solutions that can be derived for particular initial conditions or boundary values.In the context of Euler equations and the solution process we used:
- The form \( y = x^\alpha (C_1 \cos(\beta \ln x) + C_2 \sin(\beta \ln x)) \) embodies the general solution.
- The constants \( C_1 \) and \( C_2 \) allow flexibility to meet various initial or boundary requirements.
- This solution represents a family of possible solutions, not just one trajectory or behavior.
