Problem 10 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given differential equation. $$4 x^{2} y^{\prime \prime}+4 x y^{\prime}-y=0$$

Step-by-Step Solution

Answer

The solution is \( y(x) = C_1 + C_2 x^{3/4} \).

Step 1: Identify the Type of Differential Equation

The equation is a second-order linear differential equation with variable coefficients: \( 4x^2 y'' + 4xy' - y = 0 \). This equation is of Cauchy-Euler form, which is characterized by its structure compatible with the formula: \( a x^2 y'' + b x y' + c y = 0 \).

Step 2: Assume a Solution

For a Cauchy-Euler equation, we assume a solution of the form \( y = x^m \). We will find \( m \) by substituting this assumed solution into the differential equation.

Step 3: Find First and Second Derivatives

Compute the derivatives for \( y = x^m \):- \( y' = mx^{m-1} \)- \( y'' = m(m-1)x^{m-2} \). Substitute \( y \), \( y' \), and \( y'' \) back into the differential equation.

Step 4: Substitute into Differential Equation

Plug \( y = x^m \), \( y' = mx^{m-1} \), and \( y'' = m(m-1)x^{m-2} \) into the differential equation: \[ 4x^2(m(m-1)x^{m-2}) + 4x(mx^{m-1}) - x^m = 0 \].

Step 5: Simplify and Collect Like Terms

Simplify the equation: \( 4m(m-1)x^m + 4mx^m - x^m = 0 \). Collect the like terms to get: \( (4m^2 - 3m)x^m = 0 \).

Step 6: Solve the Characteristic Equation

The characteristic equation derived from the above simplification is \( 4m^2 - 3m = 0 \).Factor this equation: \( m(4m - 3) = 0 \).This yields the roots \( m = 0 \) or \( m = \frac{3}{4} \).

Step 7: General Solution

The general solution to the differential equation is derived from the roots of the characteristic equation. Since we have distinct roots, the solution is: \( y(x) = C_1 x^0 + C_2 x^{\frac{3}{4}} \). Simplifying gives \( y(x) = C_1 + C_2 x^{\frac{3}{4}} \).

Key Concepts

Second-order differential equation

A differential equation involves variables and their derivatives. When the highest derivative is the second derivative, it's called a second-order differential equation. In our example, the expression is:

• \( y'' \), which denotes the second derivative.
• \( 4x^2 y'' + 4xy' - y = 0 \) is in second-order.

The main goal is to solve for \( y \) by finding a function that satisfies this equation.
Second-order differential equations can be linear or non-linear, based on how \( y \) and its derivatives appear in the equation.
In this case, the equation is linear because it involves \( y \), \( y' \), and \( y'' \) in a linear manner.
Understanding the order helps us choose the right method for solving the equation.

Characteristic equation

The characteristic equation is a tool for solving linear differential equations with constant or even variable coefficients.
In Cauchy-Euler equations, we assume solutions of the form \( y = x^m \). By substituting this form into the original equation, we derive an algebraic equation in terms of \( m \), known as the characteristic equation.

• For our problem, the substitution results in: \( 4m^2 - 3m = 0 \).
• This equation can be solved for \( m \) to find the potential solutions for the differential equation.

Solving this characteristic equation yields the roots \( m = 0 \) and \( m = \frac{3}{4} \), which are essential in constructing the general solution.
It transforms the original complex differential problem into a more straightforward algebra problem.

Variable coefficients

In some differential equations, coefficients change with the independent variable rather than being constant. These are called differential equations with variable coefficients.
In the provided equation, the coefficients are dependent on \( x \):

• The term \( 4x^2 \) is the coefficient of the second derivative \( y'' \).
• \( 4x \) is the coefficient for the first derivative \( y' \).

This variation makes variable-coefficient equations more challenging to solve directly.
However, recognizing the Cauchy-Euler form helps us to apply a specific method, assuming a solution of \( y = x^m \).
The assumption transforms it into a characteristic equation, simplifying the process.

General solution of differential equations

The general solution of a differential equation represents all possible solutions.
For Cauchy-Euler equations like ours, once we find the roots of the characteristic equation, these roots are used to construct the general solution:

• If the roots are distinct, like \( m = 0 \) and \( m = \frac{3}{4} \), the general solution is a sum of terms, each involving \( x \) raised to one of the roots:

\[ y(x) = C_1 x^0 + C_2 x^{\frac{3}{4}} \]

• Here, \( C_1 \) and \( C_2 \) are constants that can be determined if initial conditions are provided.

This general solution is significant because it informs us about the possible family of functions that satisfy the differential equation under various conditions.