Higher Order Linear Homogeneous Differential Equations with Constant Coefficients
Solved Problems
Example 1.
Solve the differential equation \[y^{\prime\prime\prime} + 2y^{\prime\prime} - y' - 2y = 0.\]
Solution.
Write the corresponding characteristic equation:
\[{\lambda ^3} + 2{\lambda ^2} - \lambda - 2 = 0.\]
Solving it, we find the roots:
\[{\lambda ^2}\left( {\lambda + 2} \right) - \left( {\lambda + 2} \right) = 0,\;\; \Rightarrow \left( {\lambda + 2} \right)\left( {{\lambda ^2} - 1} \right) = 0,\;\; \Rightarrow \left( {\lambda + 2} \right)\left( {\lambda - 1} \right)\left( {\lambda + 1} \right) = 0,\;\; \Rightarrow {\lambda _1} = - 2,\;\; {\lambda _2} = 1,\;\; {\lambda _3} = - 1.\]
It is seen that all three roots are real. Therefore, the general solution of the differential equations can be written as
\[y\left( x \right) = {C_1}{e^{ - 2x}} + {C_2}{e^x} + {C_3}{e^{ - x}},\]
where \({C_1},\) \({C_2},\) \({C_3}\) are arbitrary constants.
Example 2.
Solve the equation \[y^{\prime\prime\prime} - 7y^{\prime\prime} + 11y' - 5y = 0.\]
Solution.
The corresponding characteristic equation is
\[{\lambda ^3} - 7{\lambda ^2} + 11\lambda - 5 = 0.\]
It is easy to see that one of the roots is the number \(\lambda = 1.\) Then, factoring the term \(\left( {\lambda - 1} \right)\) from the equation, we obtain
\[{\lambda ^3} - {\lambda ^2} - 6{\lambda ^2} + 6\lambda + 5\lambda - 5 = 0,\;\; \Rightarrow
{\lambda ^2}\left( {\lambda - 1} \right) - 6\lambda \left( {\lambda - 1} \right) + 5\left( {\lambda - 1} \right) = 0,\;\; \Rightarrow
\left( {\lambda - 1} \right) \left( {{\lambda ^2} - 6\lambda + 5} \right) = 0,\;\; \Rightarrow
\left( {\lambda - 1} \right) \left( {\lambda - 1} \right) \left( {\lambda - 5} \right) = 0,\;\; \Rightarrow
{\left( {\lambda - 1} \right)^2}\left( {\lambda - 5} \right) = 0.\]
Thus, the equation has two roots \({\lambda _1} = 1,\) \({\lambda _2} = 5,\) the first of which has multiplicity \(2.\) Then the general solution of differential equations can be written as follows:
\[y\left( x \right) = \left( {{C_1} + {C_2}x} \right){e^x} + {C_3}{e^{5x}},\]
where \({C_1},\) \({C_2},\) \({C_3}\) are arbitrary numbers.
Example 3.
Solve the equation \[y^{IV} - y^{\prime\prime\prime} + 2y' = 0.\]
Solution.
Write the characteristic equation:
\[{\lambda ^4} - {\lambda ^3} + 2\lambda = 0.\]
Factor the left side and find the roots:
\[\lambda \left( {{\lambda ^3} - {\lambda ^2} + 2} \right) = 0.\]
Note that one of the roots of the cubic polynomial is the number \(\lambda = -1.\) Therefore, we divide \({{\lambda ^3} - {\lambda ^2} + 2}\) by \(\lambda + 1:\)
\[\frac{{{\lambda ^3} - {\lambda ^2} + 2}}{{\lambda + 1}} = {\lambda ^2} - 2\lambda + 2.\]
As a result, the characteristic equation takes the following form:
\[\lambda \left( {\lambda + 1} \right) \left( {{\lambda ^2} - 2\lambda + 2} \right) = 0.\]
We find the roots of the quadratic equation:
\[{\lambda ^2} - 2\lambda + 2 = 0,\;\; \Rightarrow
D = 4 - 8 = - 4,\;\; \Rightarrow
\lambda = \frac{{2 \pm \sqrt { - 4} }}{2} = \frac{{2 \pm 2i}}{2} = 1 \pm i.\]
Thus, the characteristic equation has four distinct roots, two of which are complex:
\[{\lambda _1} = 0,\;\; {\lambda _2} = - 1,\;\; {\lambda _{3,4}} = 1 \pm i.\]
The general solution of the differential equation can be represented as
\[y\left( x \right) = {C_1} + {C_2}{e^{ - x}} + {e^x}\left( {{C_3}\cos x + {C_4}\sin x} \right),\]
where \({C_1}, \ldots, {C_4}\) are arbitrary constants.
Example 4.
Solve the equation \[{y^V} + 18y^{\prime\prime\prime} + 81y' = 0.\]
Solution.
The characteristic equation can be written as
\[{\lambda ^5} + 18{\lambda ^3} + 81\lambda = 0.\]
Factor the left side and calculate the roots:
\[\lambda \left( {{\lambda ^4} + 18{\lambda ^2} + 81} \right) = 0,\;\; \Rightarrow
\lambda {\left( {{\lambda ^2} + 9} \right)^2} = 0.\]
>As it can be seen, the equation has the following roots:
\[{\lambda _1} = 0,\;\; {\lambda _{2,3}} = \pm 3i,\]
and imaginary roots have multiplicity \(2.\) In accordance with the rules set out above, we write the general solution in the form
\[y\left( x \right) = {C_1} + \left( {{C_2} + {C_3}x} \right)\cos 3x + \left( {{C_4} + {C_5}x} \right)\sin 3x,\]
where \({C_1}, \ldots, {C_5}\) are arbitrary numbers.
Example 5.
Solve the differential equation \[y^{IV} - 4y^{\prime\prime\prime} + 5y^{\prime\prime} - 4y' + 4y = 0.\]
Solution.
Calculate the roots of the characteristic equation
\[{\lambda ^4} - 4{\lambda ^3} + 5{\lambda ^2} - 4\lambda + 4 = 0.\]
Factor the left side:
\[{\lambda ^4} - 2{\lambda ^3} - 2{\lambda ^3} + 4{\lambda ^2} + {\lambda ^2} - 2\lambda - 2\lambda + 4 = 0,\;\; \Rightarrow
\left( {{\lambda ^4} - 2{\lambda ^3}} \right) - \left( {2{\lambda ^3} - 4{\lambda ^2}} \right) + \left( {{\lambda ^2} - 2\lambda } \right) - \left( {2\lambda - 4} \right) = 0,\;\; \Rightarrow
{\lambda ^3}\left( {\lambda - 2} \right) - 2{\lambda ^2}\left( {\lambda - 2} \right) + \lambda \left( {\lambda - 2} \right) - 2\left( {\lambda - 2} \right) = 0,\;\;\Rightarrow
\left( {\lambda - 2} \right) \left( {{\lambda ^3} - 2{\lambda ^2} + \lambda - 2} \right) = 0,\;\; \Rightarrow
\left( {\lambda - 2} \right) \left[ {{\lambda ^2}\left( {\lambda - 2} \right) + \lambda - 2} \right] = 0,\;\; \Rightarrow
\left( {\lambda - 2} \right)\left( {\lambda - 2} \right) \left( {{\lambda ^2} + 1} \right) = 0,\;\;\Rightarrow
{\left( {\lambda - 2} \right)^2}\left( {{\lambda ^2} + 1} \right) = 0.\]
We see that the roots of the equation are equal
\[{\lambda _1} = 2,\;\; {\lambda _{3,4}} = \pm i.\]
The first root is of multiplicity \(2.\) The general solution of the differential equation is given by
\[y\left( x \right) = \left( {{C_1} + {C_2}x} \right){e^{2x}}
+ {C_3}\cos x + {C_4}\sin x,\]
where \({C_1}, \ldots, {C_4}\) are as usual arbitrary constants.
