Differential Equations

Second Order Equations

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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

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:

\[{k^2} - 6k + 5 = 0.\]

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.

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