# Second Order Linear Homogeneous Differential Equations with Constant Coefficients

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:

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

## Solved Problems

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### 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:

${k^2} - 6k + 5 = 0.$

Eliminate the constant $$C$$ from the system of equations:

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

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:

${k^2} - 6k + 9 = 0,\;\; \Rightarrow D = 36 - 4 \cdot 9 = 0,\;\; \Rightarrow {k_{1,2}} = 3.$

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

$y\left( x \right) = \left( {{C_1}x + {C_2}} \right){e^{3x}},$

where $${C_1},$$ $${C_2}$$ are arbitrary real numbers.

See more problems on Page 2.