Problem 10 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The general solution is \( y(x) = C_1 x^{1/2} + C_2 x^{-1/2} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( 4x^2 y'' + 4xy' - y = 0 \). This is a second-order linear homogeneous differential equation. The presence of \( x^2 \) and \( x \) with the derivatives suggests it is Cauchy-Euler type.

Step 2: Assume a Solution Form

For Cauchy-Euler equations, we assume a solution of the form \( y = x^m \), where \( m \) is a constant. This assumption is suitable because the equation's form suggests that solutions could involve powers of \( x \).

Step 3: Compute Derivatives

Calculate the first and second derivatives of the assumed solution: - The first derivative: \( y' = mx^{m-1} \).- The second derivative: \( y'' = m(m-1)x^{m-2} \).

Step 4: Substitute Derivatives into the Equation

Substitute \( y = x^m \), \( y' = mx^{m-1} \), 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 the Equation

Simplify each term:- \( 4x^2 m(m-1)x^{m-2} = 4m(m-1)x^m \),- \( 4x(mx^{m-1}) = 4mx^m \),- The equation becomes: \[ 4m(m-1)x^m + 4mx^m - x^m = 0 \]. Factor out \( x^m \):\[ x^m (4m(m-1) + 4m - 1) = 0 \].

Step 6: Solve the Characteristic Equation

Since \( x^m eq 0 \) for \( x > 0 \), solve the characteristic equation:\[ 4m(m-1) + 4m - 1 = 0 \].Distribute and combine like terms:\[ 4m^2 - 4m + 4m - 1 = 4m^2 - 1 = 0 \].Add 1 to both sides:\[ 4m^2 = 1 \].Divide by 4:\[ m^2 = \frac{1}{4} \].Take the square root:\[ m = \pm \frac{1}{2} \].

Step 7: Formulate the General Solution

The general solution for a Cauchy-Euler equation with real and distinct roots \( m_1 \) and \( m_2 \) is \( y(x) = C_1 x^{m_1} + C_2 x^{m_2} \). Substitute the values of \( m \):\[ y(x) = C_1 x^{1/2} + C_2 x^{-1/2} \].

Key Concepts

Second-order Linear Differential Equations

Second-order linear differential equations are important in many fields of science and engineering. These equations involve the second derivative of a function and can be written in the general form:

• \( a(x)y'' + b(x)y' + c(x)y = 0 \)

The coefficients \(a(x)\), \(b(x)\), and \(c(x)\) are typically functions of \(x\), and the equation does not include any term that is not multiplied by the dependent variable or its derivatives.
In this form, each term represents a different contribution to the behavior of solutions. The function \(y'\) signifies the rate of change, while \(y''\) reflects acceleration or concavity in context. Understanding how these factors interact is key to solving these equations.
This type of equation falls under the broad category of linear differential equations due to its linear, additive nature concerning the unknown function \(y\) and its derivatives.

Characteristic Equation

The characteristic equation is a crucial tool for solving linear homogeneous differential equations, particularly when the equation takes the Cauchy-Euler form. It predicts the behavior of solutions based on the assumed form of the solution.
The Cauchy-Euler equation makes it possible to assume a solution in the form \(y = x^m\), guiding to the characteristic equation after substitution and simplification. Doing so helps identify possible values for \(m\), which satisfy the differential equation.
To derive the characteristic equation for a standard Cauchy-Euler equation, you substitute \(y = x^m\) into the given equation and simplify. You arrive at a polynomial equation in terms of \(m\), called the characteristic equation. Solving this equation gives one or more roots, which in turn become exponents in the proposed solution form.
In the associated example, solving the characteristic equation \(4m^2 - 1 = 0\) yields roots \(m = \pm \frac{1}{2}\). These roots are then used to construct the general solution.

General Solution of Differential Equations

Once you have determined the roots of the characteristic equation, you can construct the general solution for the differential equation.
For a second-order linear homogeneous differential equation with distinct real roots, \(m_1\) and \(m_2\), the general solution is written as:

• \( y(x) = C_1 x^{m_1} + C_2 x^{m_2} \)

Here, \(C_1\) and \(C_2\) are arbitrary constants. These can be determined if initial or boundary conditions are provided, adding flexibility in how applications are solved.
This form of solution shows the beauty of Cauchy-Euler equations. The power functions \(x^{m_1}\) and \(x^{m_2}\) speak directly to how variables within the equation multiply and influence solutions.
In the exercise, with roots \(m = \pm \frac{1}{2}\), the general solution becomes:

• \( y(x) = C_1 x^{1/2} + C_2 x^{-1/2} \)

This captures how specific transformations of \(x\) respond in behavior, fitting precisely for the problem's structure.