91Ó°ÊÓ

When you are working with second-order linear differential equations, a **characteristic equation** is a critical concept.
It transforms the differential equation into an algebraic equation which can be solved using methods for quadratic equations.

In our exercise, the original differential equation is given by:
\[y'' + 10y' + 25y = 0\]

To form the characteristic equation, we substitute the derivatives with powers of a variable, usually denoted as \(r\).
Specifically, we replace \(y''\) with \(r^2\), \(y'\) with \(r\), and \(y\) with \(1\).
This substitution leads to:
\[r^2 + 10r + 25 = 0\]

Now we have a quadratic equation in standard form.
Solving this characteristic equation will help us find the roots, which in turn guide us to the general solution of the differential equation.

A **homogeneous differential equation** is one in which every term is a function of the dependent variable and its derivatives.
In our case, the dependent variable is \(y\) and its derivatives, \(y'\) and \(y''\).
Also, homogeneous equations do not include any isolated constant terms.

For the given equation
\[y'' + 10y' + 25y = 0\]
each term here, \(y''\), \(10y'\), and \(25y\), includes the variable \(y\) or its derivatives.
There are no additional constant terms.
Thus, this equation is homogeneous.

Homogeneous differential equations of this type are often solved by finding the characteristic equation, as detailed in the previous section.
The roots of the characteristic equation help in constructing the general solution for the given differential equation.

The **general solution** to a second-order linear homogeneous differential equation with constant coefficients is dictated by the roots of its characteristic equation.
These roots can either be real and distinct, real and repeated, or complex.
Each scenario leads to a different form of the general solution.

In our problem, the characteristic equation
\[r^2 + 10r + 25 = 0\]
has a repeated root \(r = -5\).
For repeated roots, the general solution takes the form:
\[y(t) = (C_1 + C_2 t)e^{-5t}\]
where \(C_1\) and \(C_2\) are arbitrary constants.

These constants are determined based on initial conditions provided with the problem (if any).
In the absence of initial conditions, they remain part of the general solution.
This gives a comprehensive set of functions, representing all possible solutions to the differential equation.