Consider a differential equation of type

\[{y^{\prime\prime} + py’ + qy }={ 0,}\]

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

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

\[{k^2} + pk + q = 0.\]

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

## 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 3

Solve the differential equation \(y^{\prime\prime} – 4y’ + 5y\) \( = 0.\)### Example 4

Solve the equation \(y^{\prime\prime} + 25y = 0.\)### Example 5

Solve the equation \(y^{\prime\prime} + 4iy = 0.\)### 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:

\[{k^2} – 6k + 5 = 0.\]

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:

\[{y\left( x \right) }={ {C_1}{e^x} + {C_2}{e^{5x}},}\]

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