# Differential Equations

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

## 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.

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Problem 1
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Problems 2-5