Question

Solve the following differential equations. \(y^{\prime \prime}-4 y^{\prime}+4 y=0\)

Short answer

The general solution is \(y(t) = (C_1 + C_2 t)e^{2t}\).

Step-by-step solution

Identify the characteristic equation

Rewrite the differential equation in terms of a characteristic equation. For the differential equation \(y'' - 4y' + 4y = 0\), assume a solution of the form \(y = e^{rt}\). When substituting \(y = e^{rt}\), the differential equation becomes a polynomial in terms of \(r\).

Substitute and form the characteristic equation

Substituting \(y = e^{rt}\) into the equation gives: \(r^2 e^{rt} - 4r e^{rt} + 4 e^{rt} = 0\). Factor out \(e^{rt}\): \(e^{rt} (r^2 - 4r + 4) = 0\). Since \(e^{rt}\) is never zero, we get the characteristic equation: \(r^2 - 4r + 4 = 0\).

Solve the characteristic equation

Solve \(r^2 - 4r + 4 = 0\) using the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -4\), and \(c = 4\). Substituting these, we get: \(r = \frac{4 \pm \sqrt{16 - 16}}{2} = \frac{4 \pm 0}{2} = 2\). Therefore, we have a repeated root: \(r = 2\).

Write the general solution

Since we have a repeated root \(r = 2\), the general solution for the differential equation is: \(y(t) = (C_1 + C_2 t)e^{2t}\). Here, \(C_1\) and \(C_2\) are arbitrary constants.

Key concepts

Characteristic Equation

When solving linear differential equations, the characteristic equation is a crucial part of the process.
We start by assuming a solution of the form \(y = e^{rt}\). By substituting this assumed solution into the differential equation, we convert the differential equation into a polynomial equation in terms of \(r\).
For example, if we have the differential equation \(y'' - 4y' + 4y = 0\), we assume \(y = e^{rt}\) and substitute it in:
\(r^2 e^{rt} - 4r e^{rt} + 4 e^{rt} = 0\).
We then factor out \(e^{rt}\), leading to:
\(e^{rt} (r^2 - 4r + 4) = 0\).
Since \(e^{rt}\) is never zero, we are left with the characteristic equation:
\(r^2 - 4r + 4 = 0\).
This equation helps us find the values of \(r\), which are essential for constructing the solution to the differential equation.

Repeated Roots

Repeated roots occur when the characteristic equation has the same solution more than once.
In our example, the characteristic equation \(r^2 - 4r + 4 = 0\) can be solved using the quadratic formula:
\(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Substituting \(a = 1\), \(b = -4\), and \(c = 4\), we get:
\(r = \frac{4 \pm \sqrt{16 - 16}}{2} = \frac{4 \pm 0}{2} = 2\).
This results in a repeated root \(r = 2\).
When we encounter repeated roots, the solution to the differential equation needs to be slightly modified to include the repeated nature of the root.
We thus add a term involving \(t\) (the independent variable), as this ensures we capture the repeated root's effect properly.

General Solution

The general solution of a differential equation is a linear combination of solutions that can include exponential, polynomial, and other functions.
For the case with a repeated root \(r = 2\), the general solution takes a specific form.
We need to account for the multiplicity of the root by including an additional term that grows linearly with \(t\).
The general solution, therefore, for our example \(y'' - 4y' + 4y = 0\) with repeated root \(r = 2\) is:
\(y(t) = (C_1 + C_2 t)e^{2t}\).
Here, \(C_1\) and \(C_2\) are arbitrary constants.

• The term \(C_1 e^{2t}\) is a standard solution for the root \(r = 2\).


• The term \(C_2 t e^{2t}\) accounts for the repeated root, ensuring the solution set spans all possible behaviors of the differential equation.

This combined form ensures that every possible solution of the differential equation is represented.