Linear Homogeneous Systems of Differential Equations with Constant Coefficients
Solved Problems
Example 1.
Solve the system of differential equations by elimination: \[{x'_1} = 2{x_1} + 3{x_2},\;{x'_2} = 4{x_1} - 2{x_2}.\]
Solution.
We differentiate the first equation and then substitute the derivative \({x'_2}\) from the second equation:
\[{x^{\prime\prime}_1} = 2{x'_1} + 3{x'_2},\;\; \Rightarrow
{x^{\prime\prime}_1} = 2{x'_1} + 3\left( {4{x_1} - 2{x_2}} \right),\;\; \Rightarrow
{x^{\prime\prime}_1} = 2{x'_1} + 12{x_1} - 6{x_2}.\]
Express \(3{x_2}\) from the first equation:
\[3{x_2} = {x'_1} - 2{x_1}.\]
Substituting this into the last equation, we get:
\[{x^{\prime\prime}_1} = 2{x'_1} + 12{x_1} - 2\left( {{x'_1} - 2{x_1}} \right),\;\; \Rightarrow
{x^{\prime\prime}_1} = \cancel{2{x'_1}} + 12{x_1} - \cancel{2{x'_1}} + 4{x_1},\;\; \Rightarrow
{x^{\prime\prime}_1} - 16{x_1} = 0.\]
Find the roots of the characteristic equation:
\[{\lambda ^2} - 16 = 0,\;\; \Rightarrow {\lambda _{1,2}} = \pm 4.\]
Hence, the general solution of the \(2\)nd order equation for the variable \({x_1}\) is given by
\[{x_1}\left( t \right) = {C_1}{e^{4t}} + {C_2}{e^{ - 4t}},\]
where \({C_1},\) \({C_2}\) are arbitrary constants.
Now we compute the derivative \({x'_1}\) and substitute the expressions for \({x_1},\) \({x'_1}\) in the first equation of the original system:
\[{x'_1}\left( t \right) = 4{C_1}{e^{4t}} - 4{C_2}{e^{ - 4t}},\;\; \Rightarrow
4{C_1}{e^{4t}} - 4{C_2}{e^{ - 4t}} = 2{C_1}{e^{4t}} + 2{C_2}{e^{ - 4t}} + 3{x_2},\;\; \Rightarrow
3{x_2} = 2{C_1}{e^{4t}} - 6{C_2}{e^{ - 4t}},\;\; \Rightarrow
{x_2} = \frac{2}{3}{C_1}{e^{4t}} - 2{C_2}{e^{ - 4t}}.\]
To keep integer coefficients, it is convenient to designate: \({C_1} \to 3{C_1}.\) As a result, we obtain the final solution in the following form:
\[\left\{ \begin{array}{l}
{x_1}\left( t \right) = 3{C_1}{e^{4t}} + {C_2}{e^{ - 4t}}\\
{x_2}\left( t \right) = 2{C_1}{e^{4t}} - 2{C_2}{e^{ - 4t}}
\end{array} \right..\]
Example 2.
Solve the system by elimination: \[x' = 6x - y,\;y' = x + 4y.\]
Solution.
We convert this system to a single \(2\)nd order equation for the function \(x\left( t \right).\) Differentiating the first equation and substituting \(y'\) from the second equation, we have:
\[x^{\prime\prime} = 6x' - y',\;\; \Rightarrow
x^{\prime\prime} = 6x' - \left( {x + 4y} \right),\;\; \Rightarrow
x^{\prime\prime} = 6x' - x - 4y.\]
Express the variable \(y\) in terms of \(x\) and \(x'\) from the first equation:
\[y = 6x - x',\;\; \Rightarrow
x^{\prime\prime} = 6x' - x - 4\left( {6x - x'} \right),\;\; \Rightarrow
x^{\prime\prime} = 6x' - x - 24x + 4x',\;\; \Rightarrow
x^{\prime\prime} - 10x' + 25x = 0.\]
Compute the roots of the auxiliary equation:
\[{\lambda ^2} - 10\lambda + 25 = 0,\;\; D = 0,\;\; \Rightarrow
{\lambda _1} = 5.\]
So, we have one root \(\lambda = 5\) of multiplicity \(2.\) Consequently, the general solution for the function \(x\left( t \right)\) is written as
\[x\left( t \right) = \left( {{C_1} + {C_2}t} \right){e^{5t}},\]
where \({C_1},\) \({C_2}\) are arbitrary numbers.
Find the derivative \(x'\left( t \right)\) and substituting it in the first equation of the original system determine the function \(y\left( t \right):\)
\[x'\left( t \right) = {C_2}{e^{5t}} + \left( {5{C_1} + 5{C_2}t} \right){e^{5t}}
= \left( {5{C_1} + {C_2} + 5{C_2}t} \right){e^{5t}},\;\; \Rightarrow
\left( {5{C_1} + {C_2} + 5{C_2}t} \right){e^{5t}} = \left( {6{C_1} + 6{C_2}t} \right){e^{5t}} - y,\;\; \Rightarrow
y = \left( {{C_1} - {C_2} + {C_2}t} \right){e^{5t}}.\]
Hence, the general solution is written as
\[\left\{ \begin{array}{l}
x\left( t \right) = \left( {{C_1} + {C_2}t} \right){e^{5t}}\\
y\left( t \right) = \left( {{C_1} - {C_2} + {C_2}t} \right){e^{5t}}
\end{array} \right..\]
Example 3.
Find the general solution of the system \[{x'_1} = 5{x_1} + 2{x_2},\;{x'_2} = - 4{x_1} + {x_2}.\]
Solution.
Differentiating the first equation, we get:
\[{x^{\prime\prime}_1} = 5{x'_1} + 2{x'_2}.\]
Substitute the derivative \({x'_2}\) from the second equation:
\[{x^{\prime\prime}_1} = 5{x'_1} + 2\left( { - 4{x_1} + {x_2}} \right),\;\; \Rightarrow {x^{\prime\prime}_1} = 5{x'_1} - 8{x_1} + 2{x_2}.\]
Express \(2{x_2}\) in terms of \({x_1}\) from the first equation:
\[{x^{\prime\prime}_1} = 5{x'_1} - 8{x_1} + {x'_1} - 5{x_1},\;\; \Rightarrow
{x^{\prime\prime}_1} - 6{x'_1} + 13{x_1} = 0.\]
We have obtained a homogeneous equation of the \(2\)nd order with constant coefficients. As usual, we construct the general solution using the characteristic equation:
\[{\lambda ^2} - 6\lambda + 13 = 0,\;\; D = 36 - 52 = - 16,\;\; \Rightarrow
{\lambda _{1,2}} = \frac{{6 \pm \sqrt { - 16} }}{2}
= \frac{{6 \pm 4i}}{2} = 3 \pm 2i.\]
As it can be seen, the auxiliary equation has roots in the form of a pair of complex conjugate numbers. The general solution for the function \({x_1}\left( t \right)\) can be written as
\[{x_1}\left( t \right) = {e^{3t}}\left( {{C_1}\cos 2t + {C_2}\sin 2t} \right),\]
where \({C_1},\) \({C_2}\) are arbitrary constants.
Now we find another function \({x_1}\left( t \right).\) The derivative \({x'_1}\) is
\[{x'_1}\left( t \right) = 3{e^{3t}}\left( {{C_1}\cos 2t + {C_2}\sin 2t} \right)
+ {e^{3t}}\left( { - 2{C_1}\sin 2t + 2{C_2}\cos 2t} \right)
= {e^{3t}}\left[ {\left( {3{C_1} + 2{C_2}} \right)\cos 2t + \left( {3{C_2} - 2{C_1}} \right)\sin 2t} \right].\]
Substituting \({x_1}\) and \({x'_1}\) in the first equation, we get:
\[{e^{3t}}\left[ {\left( {3{C_1} + 2{C_2}} \right)\cos 2t + \left( {3{C_2} - 2{C_1}} \right)\sin 2t} \right]
= 5{e^{3t}}\left( {{C_1}\cos 2t + {C_2}\sin 2t} \right) + 2{x_2},\;\; \Rightarrow
2{x_2} = {e^{3t}}\left[ {\left( {2{C_2} - 2{C_1}} \right)\cos 2t - \left( {2{C_2} + 2{C_1}} \right)\sin 2t} \right],\;\; \Rightarrow
{x_2} = {e^{3t}}\left[ {\left( {{C_2} - {C_1}} \right)\cos 2t - \left( {{C_2} + {C_1}} \right)\sin 2t} \right].\]
So, the general solution is given by
\[\left\{ \begin{array}{l}
{x_1}\left( t \right) = {e^{3t}}\left[ {{C_1}\cos 2t + {C_2}\sin 2t} \right]\\
{x_2}\left( t \right) = {e^{3t}}\left[ {\left( {{C_2} - {C_1}} \right)\cos 2t - \left( {{C_2} + {C_1}} \right)\sin 2t} \right]
\end{array} \right..\]
