Problem 10 - Technical Mathematics with Calculus

Question

Find the general solution to each differential equation. $$y^{\prime \prime}-4 y^{\prime}+y=0$$

Step-by-Step Solution

Answer

The general solution of the differential equation \(y'' - 4y' + 4y = 0\) is \(y(t) = C_1 e^{2t} + C_2 t e^{2t}\), where \(C_1\) and \(C_2\) are constants.

Step 1: Find the characteristic equation

To solve the second-order linear homogeneous differential equation, we need to find its characteristic equation. This is done by replacing the derivatives in the equation with powers of a variable, typically 'r'. For the equation given, replace the second derivative term with \(r^2\), the first derivative with \(r\), and the zero-order term with 1. The characteristic equation will therefore be \(r^2 - 4r + 4 = 0\).

Step 2: Solve the characteristic equation

Solve the characteristic equation \(r^2 - 4r + 4 = 0\) by factoring or using the quadratic formula. The equation factors as \((r - 2)^2 = 0\), which means there is a repeated root at \(r = 2\).

Step 3: Write the general solution

Since we have a repeated root, the general solution to the differential equation will be of the form \(y(t) = C_1 e^{rt} + C_2 t e^{rt}\), where \(r = 2\), \(C_1\) and \(C_2\) are constants determined by initial conditions. Therefore, the general solution is \(y(t) = C_1 e^{2t} + C_2 t e^{2t}\).

Key Concepts

Characteristic Equation

Solving a second-order linear homogeneous differential equation begins with finding the characteristic equation. This crucial step involves a clever transformation where we convert the differential equation into a polynomial format that is much easier to solve. Imagine replacing the calculus terms with algebraic terms, which is exactly what occurs. In our specific exercise, we replace the second derivative term, denoted as \(y''\), with \(r^2\), and for the first derivative term, \(y'\), we use simply \(r\). The constant term remains unchanged.

Deriving the characteristic equation from \(y'' - 4y' + y = 0\) results in the polynomial \(r^2 - 4r + 4 = 0\). It’s a transformation that turns the original, possibly intimidating differential equation into a familiar quadratic equation, thus bridging the gap between two major areas in mathematics: calculus and algebra.

Differential Equation Solution

Now that we have the characteristic equation, the next step is solving it. This is where the quadratic formula, factoring, or possibly completing the square comes into play. We're hunting for the roots, which in the context of differential equations, will guide us to the solution.

In our example, the characteristic equation factors nicely into \((r - 2)^2 = 0\), indicating a repeated root at \(r = 2\). If we had distinct roots, each would inform a particular part of the solution. However, with repeated roots, the solution takes a particular form that includes the exponential function, ensuring that the calculus characteristics of the original equation aren't lost in our algebraic translation.

General Solution of Differential Equation

With the roots in hand, we are ready to write down the general solution of the differential equation. This solution encapsulates all possible functions that satisfy the original equation. In cases of repeated roots, like in our exercise where \(r = 2\), the general solution combines exponential terms and polynomial terms to capture the behavior dictated by the differential equation.

To be precise, the solution is represented as \(y(t) = C_1 e^{rt} + C_2 t e^{rt}\), where \(C_1\) and \(C_2\) are constants that are typically determined by initial value conditions. For our example, this translates to \(y(t) = C_1 e^{2t} + C_2 t e^{2t}\). Here, the term \(t e^{2t}\) is included specifically because of the repeated root, showcasing how the nature of the roots influences the form of the general solution.