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Magazine Chapter 2: Problem 2
Find a real general solution, showing the details of your work. $$4 x^{2} y^{n}+4 x y^{\prime}-y=0$$
Short Answer
Expert verified
The general solution is \(y = Cx^{1/4}\), with \(C\) as a constant.
Step by step solution
01
Recognize the Form
The given differential equation is \(4x^2 y^n + 4x y' - y = 0\). This resembles the form of an Euler-Cauchy differential equation, which is usually expressed as \(ax^2 y'' + bxy' + cy = 0\). However, here, we are dealing with a function of another form where separation of variables could be a possible method.
02
Assume a Power Function
We consider the substitution \(y = x^m\) to ascertain the values of \(m\) for which the equation holds. We will then substitute \(y = x^m\) and \(y' = mx^{m-1}\) into the equation.
03
Substitute y and y'
Substitute \(y = x^m\) and \(y' = mx^{m-1}\) into the equation, yielding: \[4x^2 (x^m)^n + 4x(mx^{m-1}) - x^m = 0\] which simplifies to \[4x^{2 + mn} + 4mx^m - x^m = 0.\]
04
Group Like Terms
The expression is now \(4x^{2 + mn} + (4m - 1)x^m = 0\). For this to be true for all \(x\), both terms must independently equal zero.
05
Solve for Exponent m
Since each term must be zero separately:1. \(4x^{2 + mn} = 0\) implies \(mn = -2\);2. \(4m - 1 = 0\) implies \(m = \frac{1}{4}\).Substitute \(m = \frac{1}{4}\) into \(mn = -2\) to find \(n\). Thus, \(n = -2 \times 4 = -8\).
06
Write the General Solution
Since \(y = x^m\) and \(m = \frac{1}{4}\), the function has a form \(y = Cx^{1/4}\), where \(C\) is a general constant that can be determined by initial conditions, if any.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separation of Variables
Separation of variables is a useful technique for solving differential equations, often when the equation can be rearranged so that each variable appears on a different side of the equation. This method is particularly helpful for first-order differential equations where you can isolate the variables and integrate each side separately.
Here's how it typically works:
Here's how it typically works:
- First, try to express the equation in a form where all terms involving one variable and its derivatives are on one side of the equation, and terms involving the other variable are on the other side.
- Next, divide both sides of the equation by respective variables or functions to truly "separate" the variables.
- Finally, integrate both sides separately to solve the equation.
Power Function Substitution
Power function substitution is a clever technique used to transform a differential equation into a simpler form. This is especially useful for equations like the Euler-Cauchy type, where exponents naturally appear in the solution. In our exercise, the substitution \(y = x^m\) was employed.
The steps involved in power function substitution include:
The steps involved in power function substitution include:
- Choosing an appropriate form of substitution, usually \(y = x^m\), to simplify the given differential equation by reducing its complexity.
- Substituting \(y\) and its derivative \(y' = mx^{m-1}\) back into the equation to transform it.
- Simplifying the expression to group like terms, which helps isolate powers or constants associated with the variables.
Differential Equations Solutions
Solving differential equations involves finding a function or set of functions that satisfy the equation. The original task was to solve an equation of a type resembling Euler-Cauchy differential equations, which are characterized by their polynomial-like structure involving both the function and its derivatives.
To solve such equations:
To solve such equations:
- Recognize the equation's form and determine if it can be simplified. For instance, the exercise showed a non-standard format, hinting at the usefulness of substitution techniques.
- Once simplified, determine the general solution. In our example, the power function substitution allowed simplification, resulting in the general solution \(y = Cx^{1/4}\), where \(C\) is an arbitrary constant.
- Explore the solution further by integrating, when necessary, or applying initial conditions to determine the constant of integration \(C\).
