Problem 11 - Technical Mathematics with Calculus

Question

Second-Order DE, Roots of Auxiliary Equation Real and Equal $$y^{\prime \prime}-4 y^{\prime}+4 y=0$$

Step-by-Step Solution

Answer

The solution to the differential equation \(y^{\text{''}} - 4y' + 4y = 0\) is \(y = (C_1 + C_2x)e^{2x}\) where \(C_1\) and \(C_2\) are constants determined by initial conditions.

Step 1: Write down the auxiliary equation

Convert the given second-order differential equation into its characteristic auxiliary equation by replacing the second, first, and zeroth derivatives with the corresponding powers of a variable, often denoted as 'm'. For the DE, our auxiliary equation is: \(m^2 - 4m + 4 = 0 \).

Step 2: Solve the auxiliary equation

Solve the auxiliary equation for its roots. The quadratic equation \(m^2 - 4m + 4 = 0\) can be factored as \((m - 2)^2 = 0\), implying a repeated root of \(m = 2\).

Step 3: Write down the general solution for repeated real roots

For a second-order differential equation with repeated real roots, the general solution is of the form \(y = (C_1 + C_2x)e^{mx}\), where \(C_1\) and \(C_2\) are constants to be determined from initial conditions and \(m\) is the repeated root. Since \(m = 2\), the solution becomes \(y = (C_1 + C_2x)e^{2x}\).

Key Concepts

Auxiliary Equation

When tackling a second-order differential equation (DE), it's crucial to translate the problem into something more manageable. The auxiliary equation, also known as the characteristic equation, serves as a bridge between differential equations and algebra. In the given exercise, the DE \( y^{\prime \prime}-4 y^{\prime}+4 y=0 \) is converted into its auxiliary form by substituting \( y^{\prime \prime} \) with \( m^2 \) , \( y^{\prime} \) with \( m \) and \( y \) itself is treated as a constant.

The resulting auxiliary equation, \( m^2 - 4m + 4 = 0 \) , is a quadratic equation which allows us to find the roots that are crucial for solving the DE. By using auxiliary equations, we turn the problem from calculus into one that can be approached with the tools of basic algebra.

Characteristic Roots

Characteristic roots, also known as eigenvalues or solution of the auxiliary equation, play a pivotal role in determining the general solution of a DE. They are obtained by solving the auxiliary equation. In our case, the auxiliary quadratic equation \( m^2 - 4m + 4 = 0 \) can be factored easily which reveals a repeated root of \( m=2 \) .

The nature of these roots—whether they are real or complex, distinct or repeated—directly influences the form of the general solution. Repeated real roots, which we have here, imply a specific solution structure involving both exponential and linear terms multiplied together, ensuring the solution encompasses all possible behaviors of the original DE.

General Solution of DE

The general solution of a differential equation incorporates all possible solutions of the DE, encompassing any constants that would be modified to fit specific boundary or initial conditions. When we have repeated real roots, as identified in our exercise with the root \( m=2 \), the general solution takes on a particular format.

In this case, the solution is an exponential function affected by the characteristic root, which is also multiplied by a linear expression to account for the multiplicity of the root: \( y = (C_1 + C_2x)e^{mx} \) . Here, \( C_1 \) and \( C_2 \) are constants determined by the initial conditions of the DE, and \( m \) is the repeated root. With \( m=2 \) determined from our auxiliary equation, the final form of the general solution for the given DE is \( y = (C_1 + C_2x)e^{2x} \) . Noting that for different types of roots, such as distinct real roots or complex roots, the general solution would have a different structure.