Consider a differential equation of type

where \(p, q\) are some constant coefficients.

For each of the equation we can write the so-called characteristic (auxiliary) equation:

The general solution of the homogeneous differential equation depends on the roots of the characteristic quadratic equation. There are the following options:

1. Discriminant of the characteristic quadratic equation \(D \gt 0.\) Then the roots of the characteristic equations \({k_1}\) and \({k_2}\) are real and distinct. In this case the general solution is given by the following function
   \[{y\left( x \right) }={ {C_1}{e^{{k_1}x}} + {C_2}{e^{{k_2}x}},}\]

   where \({C_1}\) and \({C_2}\) are arbitrary real numbers.

2. Discriminant of the characteristic quadratic equation \(D = 0.\) Then the roots are real and equal. It is said in this case that there exists one repeated root \({k_1}\) of order 2. The general solution of the differential equation has the form:
   \[{y\left( x \right) }={ \left( {{C_1}x + {C_2}} \right){e^{{k_1}x}}.}\]
3. Discriminant of the characteristic quadratic equation \(D \lt 0.\) Such an equation has complex roots \({k_1} = \alpha + \beta i,\) \({k_2} = \alpha – \beta i.\) The general solution is written as
   \[{y\left( x \right) }={ {e^{\alpha x}}\left[ {{C_1}\cos \left( {\beta x} \right) }\right.}+{\left.{ {C_2}\sin \left( {\beta x} \right)} \right].}\]

Example 1.

Solve the differential equation \(y^{\prime\prime} – 6y’ + 5y = 0.\)

Solution.

First we write the corresponding characteristic equation for the given differential equation:

The roots of this equation are \({k_1} = 1,\) \({k_2} = 5.\) Since the roots are real and distinct, the general solution has the form:

where \({C_1}\) and \({C_2}\) are arbitrary constants.