# Second Order Linear Homogeneous Differential Equations with Constant Coefficients

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:

- 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.
- 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:
\[y\left( x \right) = \left( {{C_1}x + {C_2}} \right){e^{{k_1}x}}.\]
- 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
\[y\left( x \right) = {e^{\alpha x}}\left[ {{C_1}\cos \left( {\beta x} \right) + {C_2}\sin \left( {\beta x} \right)} \right].\]

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Solve the differential equation \[y^{\prime\prime} - 6y' + 5y = 0.\]

### Example 2

Find the general solution of the equation \[y^{\prime\prime} - 6y' + 9y = 0.\]

### 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:

Eliminate the constant \(C\) from the system of equations:

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

### Example 2.

Find the general solution of the equation \[y^{\prime\prime} - 6y' + 9y = 0.\]

Solution.

We write the characteristic equation and calculate its roots:

As it can be seen, the characteristic equation has one root of order \(2:\) \({k_1} = 3.\) Therefore, the general solution of the differential equation is given by

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