Linear Systems of Differential Equations with Variable Coefficients
Solved Problems
Example 1.
Write the linear system of equations with the following solutions:
\[{\mathbf{x}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
2\\
t
\end{array}} \right],\;
{\mathbf{x}_2}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
t\\
{{t^2}}
\end{array}} \right],\;
t \ne 0.\]
Solution.
In this problem the fundamental matrix of the system is known:
\[\Phi \left( t \right) = \left[ {\begin{array}{*{20}{c}}
2&t\\
t&{{t^2}}
\end{array}} \right].\]
We compute the inverse matrix \({\Phi ^{ - 1}}\left( t \right):\)
\[\Delta \left( \Phi \right) = \left| {\begin{array}{*{20}{c}}
2&t\\
t&{{t^2}}
\end{array}} \right| = 2{t^2} - {t^2} = {t^2},\;\; \Rightarrow
{\Phi ^{ - 1}}\left( t \right) = \frac{1}{{\Delta \left( \Phi \right)}}C_{ij}^T
= \frac{1}{{{t^2}}}{\left[ {\begin{array}{*{20}{c}}
{{t^2}}&{ - t}\\
{ - t}&2
\end{array}} \right]^T}
= \frac{1}{{{t^2}}}\left[ {\begin{array}{*{20}{c}}
{{t^2}}&{ - t}\\
{ - t}&2
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
1&{ - \frac{1}{t}}\\
{ - \frac{1}{t}}&{\frac{2}{{{t^2}}}}
\end{array}} \right].\]
Here \({C_{ij}}\) denote the cofactors of the corresponding elements of the fundamental matrix \(\Phi \left( t \right).\)
The coefficient matrix of the system of equations is given by
\[A\left( t \right) = \Phi'\left( t \right){\Phi ^{ - 1}}\left( t \right).\]
The derivative of the fundamental matrix (it is calculated element by element) is equal to
\[{\Phi'}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
0&1\\
1&{{2t}}
\end{array}} \right].\]
Hence, we obtain:
\[
A\left( t \right) = \left[ {\begin{array}{*{20}{c}}
0&1\\
1&{2t}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{ - \frac{1}{t}}\\
{ - \frac{1}{t}}&{\frac{2}{{{t^2}}}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{0 - \frac{1}{t}}&{0 + \frac{2}{{{t^2}}}}\\
{1 - 2}&{ - \frac{1}{t} + \frac{4}{t}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{ - \frac{1}{t}}&{\frac{2}{{{t^2}}}}\\
{ - 1}&{\frac{3}{t}}
\end{array}} \right].\]
Thus, the system of equations whose solutions are \({\mathbf{x}_1}\left( t \right),\) \({\mathbf{x}_2}\left( t \right),\) can be written as
\[\frac{{dx}}{{dt}} = - \frac{x}{t} + \frac{{2y}}{{{t^2}}},\;\; \frac{{dy}}{{dt}} = - x + \frac{{3y}}{t}.\]
Example 2.
Find a fundamental matrix of the system of differential equations
\[\frac{{dx}}{{dt}} = x + ty,\; \frac{{dy}}{{dt}} = tx + y,\]
making sure that the coefficient matrix \(A\left( t \right)\) commutes with its integral.
Solution.
We first show that the multiplication of the matrix \(A\left( t \right)\) by its integral is commutative. The original matrix is
\[A\left( t \right) = \left[ {\begin{array}{*{20}{c}}
1&t\\
t&1
\end{array}} \right].\]
The integral of the matrix \(A\left( t \right)\) is found by elementwise integration. For simplicity, we take the lower boundary of integration to be zero. Then
\[\int\limits_0^t {A\left( \tau \right)d\tau } = \left[ {\begin{array}{*{20}{c}}
t&{\frac{{{t^2}}}{2}}\\
{\frac{{{t^2}}}{2}}&t
\end{array}} \right].\]
As a result, we have
\[A\left( t \right) \cdot \int\limits_0^t {A\left( \tau \right)d\tau }
= \left[ {\begin{array}{*{20}{c}}
1&t\\
t&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
t&{\frac{{{t^2}}}{2}}\\
{\frac{{{t^2}}}{2}}&t
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{t + \frac{{{t^3}}}{2}}&{\frac{{{t^2}}}{2} + {t^2}}\\
{{t^2} + \frac{{{t^2}}}{2}}&{\frac{{{t^3}}}{2} + t}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{t + \frac{{{t^3}}}{2}}&{\frac{{3{t^2}}}{2}}\\
{\frac{{3{t^2}}}{2}}&{t + \frac{{{t^3}}}{2}}
\end{array}} \right],\]
\[\int\limits_0^t {A\left( \tau \right)d\tau } \cdot A\left( t \right)
= \left[ {\begin{array}{*{20}{c}}
t&{\frac{{{t^2}}}{2}}\\
{\frac{{{t^2}}}{2}}&t
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&t\\
t&1
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{t + \frac{{{t^3}}}{2}}&{{t^2} + \frac{{{t^2}}}{2}}\\
{\frac{{{t^2}}}{2} + {t^2}}&{\frac{{{t^3}}}{2} + t}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{t + \frac{{{t^3}}}{2}}&{\frac{{3{t^2}}}{2}}\\
{\frac{{3{t^2}}}{2}}&{t + \frac{{{t^3}}}{2}}
\end{array}} \right].
\]
So, the commutative property of the matrix product is true. Therefore, the fundamental matrix is given by
\[\Phi \left( t \right) = {e^{\,\int\limits_0^t {A\left( \tau \right)d\tau } }}
= {e^{\left[ {\begin{array}{*{20}{c}}
t&{\frac{{{t^2}}}{2}}\\
{\frac{{{t^2}}}{2}}&t
\end{array}} \right]}}.\]
We compute the matrix exponential by converting the matrix to diagonal form. In this case, the eigenvalues depend on the variable \(t\) and can be expressed as follows:
\[\left| {\begin{array}{*{20}{c}}
{t - \lambda }&{\frac{{{t^2}}}{2}}\\
{\frac{{{t^2}}}{2}}&{t - \lambda }
\end{array}} \right| = 0,\;\; \Rightarrow
{\left( {t - \lambda } \right)^2} - {\left( {\frac{{{t^2}}}{2}} \right)^2} = 0,\;\; \Rightarrow
\left| {\lambda - t} \right| = \pm \frac{{{t^2}}}{2},\;\; \Rightarrow
{\lambda _{1,2}} = t \pm \frac{{{t^2}}}{2}.\]
For each eigenvalue, we find the corresponding eigenvector. For \({\lambda _1}\) we obtain:
\[{\lambda _1} = t + \frac{{{t^2}}}{2},\;\; \Rightarrow
\left( {A - {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
{t - \left( {t + \frac{{{t^2}}}{2}} \right)}&{\frac{{{t^2}}}{2}}\\
{\frac{{{t^2}}}{2}}&{t - \left( {t + \frac{{{t^2}}}{2}} \right)}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
- \frac{{{t^2}}}{2}{V_{11}} + \frac{{{t^2}}}{2}{V_{21}} = 0,\;\; \Rightarrow
{\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right].\]
Similarly, we find the eigenvector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T}\) for the eigenvalue \({\lambda _2}:\)
\[{\lambda _2} = t - \frac{{{t^2}}}{2},\;\; \Rightarrow
\left( {A - {\lambda _2}I} \right){\mathbf{V}_2} = \mathbf{0},\;\Rightarrow
\left[ {\begin{array}{*{20}{c}}
{t - \left( {t - \frac{{{t^2}}}{2}} \right)}&{\frac{{{t^2}}}{2}}\\
{\frac{{{t^2}}}{2}}&{t - \left( {t - \frac{{{t^2}}}{2}} \right)}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
\frac{{{t^2}}}{2}{V_{12}} + \frac{{{t^2}}}{2}{V_{22}} = 0,\;\; \Rightarrow
{\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{-1}\\
1
\end{array}} \right].\]
Then the matrix of reduction to diagonal form (more precisely to Jordan form) is given by
\[H = \left[ {\begin{array}{*{20}{c}}
1&{ - 1}\\
1&1
\end{array}} \right].\]
We compute the inverse matrix \({H^{ - 1}}:\)
\[\Delta \left( H \right) = \left| {\begin{array}{*{20}{r}}
1&{ - 1}\\
1&1
\end{array}} \right| = 1 + 1 = 2,\;\; \Rightarrow
{H^{ - 1}} = \frac{1}{{\Delta \left( H \right)}}H_{ij}^T
= \frac{1}{2}{\left[ {\begin{array}{*{20}{r}}
1&{ - 1}\\
1&1
\end{array}} \right]^T}
= \frac{1}{2}\left[ {\begin{array}{*{20}{r}}
1&1\\
{ - 1}&1
\end{array}} \right].\]
Hence, the Jordan form \(J\) is as follows:
\[J = {H^{ - 1}}\left[ {\int\limits_0^t {A\left( \tau \right)d\tau } } \right]H
= \frac{1}{2}\left[ {\begin{array}{*{20}{r}}
1&1\\
{ - 1}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
t&{\frac{{{t^2}}}{2}}\\
{\frac{{{t^2}}}{2}}&t
\end{array}} \right] \left[ {\begin{array}{*{20}{r}}
1&{ - 1}\\
1&1
\end{array}} \right]
= \frac{1}{2}\left[ {\begin{array}{*{20}{c}}
{t + \frac{{{t^2}}}{2}}&{\frac{{{t^2}}}{2} + t}\\
{ - t + \frac{{{t^2}}}{2}}&{ - \frac{{{t^2}}}{2} + t}
\end{array}} \right] \left[ {\begin{array}{*{20}{r}}
1&{ - 1}\\
1&1
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
\color{blue}{t + \frac{{{t^2}}}{2}}&0\\
0&\color{red}{t - \frac{{{t^2}}}{2}}
\end{array}} \right].\]
The exponential of the matrix \(J\) is given by
\[
{e^J} = \left[ {\begin{array}{*{20}{c}}
{{e^{t + \frac{{{t^2}}}{2}}}}&0\\
0&{{e^{t - \frac{{{t^2}}}{2}}}}
\end{array}} \right]
= {e^t}\left[ {\begin{array}{*{20}{c}}
{{e^{\frac{{{t^2}}}{2}}}}&0\\
0&{{e^{ - \frac{{{t^2}}}{2}}}}
\end{array}} \right].\]
We can now calculate the fundamental matrix \(\Phi \left( t \right):\)
\[\Phi \left( t \right) = {e^{\,\int\limits_0^t {A\left( \tau \right)d\tau } }} = H{e^J}{H^{ - 1}}
= \left[ {\begin{array}{*{20}{r}}
1&{ - 1}\\
1&1
\end{array}} \right] \cdot {e^t}\left[ {\begin{array}{*{20}{c}}
{{e^{\frac{{{t^2}}}{2}}}}&0\\
0&{{e^{ - \frac{{{t^2}}}{2}}}}
\end{array}} \right] \cdot \frac{1}{2}\left[ {\begin{array}{*{20}{r}}
1&1\\
{ - 1}&1
\end{array}} \right]
= \frac{{{e^t}}}{2}\left[ {\begin{array}{*{20}{c}}
{{e^{\frac{{{t^2}}}{2}}} + 0}&{0 - {e^{ - \frac{{{t^2}}}{2}}}}\\
{{e^{\frac{{{t^2}}}{2}}} + 0}&{0 + {e^{ - \frac{{{t^2}}}{2}}}}
\end{array}} \right] \left[ {\begin{array}{*{20}{r}}
1&1\\
{ - 1}&1
\end{array}} \right]
= \frac{{{e^t}}}{2}\left[ {\begin{array}{*{20}{c}}
{{e^{\frac{{{t^2}}}{2}}}}&{ - {e^{ - \frac{{{t^2}}}{2}}}}\\
{{e^{\frac{{{t^2}}}{2}}}}&{{e^{ - \frac{{{t^2}}}{2}}}}
\end{array}} \right] \left[ {\begin{array}{*{20}{r}}
1&1\\
{ - 1}&1
\end{array}} \right]
= \frac{{{e^t}}}{2}\left[ {\begin{array}{*{20}{c}}
{{e^{\frac{{{t^2}}}{2}}} + {e^{ - \frac{{{t^2}}}{2}}}}&{{e^{\frac{{{t^2}}}{2}}} - {e^{ - \frac{{{t^2}}}{2}}}}\\
{{e^{\frac{{{t^2}}}{2}}} - {e^{ - \frac{{{t^2}}}{2}}}}&{{e^{\frac{{{t^2}}}{2}}} + {e^{ - \frac{{{t^2}}}{2}}}}
\end{array}} \right]
= {e^t}\left[ {\begin{array}{*{20}{c}}
{\cosh \frac{{{t^2}}}{2}}&{\sinh\frac{{{t^2}}}{2}}\\
{\sinh\frac{{{t^2}}}{2}}&{\cosh \frac{{{t^2}}}{2}}
\end{array}} \right].\]
Example 3.
Find the general solution of the system
\[\frac{{dx}}{{dt}} = - tx + y,\;
\frac{{dy}}{{dt}} = \left( {1 - {t^2}} \right)x + ty,\;
x \gt 0,\]
if one solution is known: \[{\mathbf{X}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{y_1}\left( t \right)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
t
\end{array}} \right].\]
Solution.
Let the second linearly independent solution be expressed by the vector function
\[{\mathbf{X}_2}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_2}\left( t \right)}\\
{{y_2}\left( t \right)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
u\\
v
\end{array}} \right]\]
with the initial condition \(u\left( {t = 0} \right) = 0,\) \(v\left( {t = 0} \right) = 1.\)
We then use Liouville's formula, which is written as
\[W\left( t \right) = \left| {\begin{array}{*{20}{c}}
1&u\\
t&v
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right|{e^{\int\limits_0^t {A\left( \tau \right)d\tau } }} = 1 \cdot {e^{\int\limits_0^t {\left( { - \tau + \tau } \right)d\tau } }} = {e^{\int\limits_0^t {0d\tau } }} = {e^0} = 1.\]
Hence we obtain the relation between the unknown functions \(u\) and \(v:\)
\[v - tu = 1.\]
Consider the second equation of the original system. Substituting the solution \({\mathbf{X}_2}\left( t \right),\) we write it in the form of
\[\frac{{dv}}{{dt}} = \left( {1 - {t^2}} \right)u + tv.\]
From the previous equation, we can express the term \(tv:\)
\[v - tu = 1,\;\; \Rightarrow tv - {t^2}u = t,\;\; \Rightarrow tv = {t^2}u + t.\]
Substitute it into the differential equation for the function \(v\left( t \right):\)
\[\frac{{dv}}{{dt}} = \left( {1 - {t^2}} \right)u + tv,\;\; \Rightarrow
\frac{{dv}}{{dt}} = \left( {1 - {t^2}} \right)u + {t^2}u + t,\;\; \Rightarrow
\frac{{dv}}{{dt}} = u - \cancel{{t^2}u} + \cancel{{t^2}u} + t,\;\; \Rightarrow
\frac{{dv}}{{dt}} = u + t,\;\; \Rightarrow
t\frac{{dv}}{{dt}} = tu + {t^2}.\]
Given that \(tu = v - 1,\) we obtain a first order linear equation for the function \(v\left( t \right):\)
\[t\frac{{dv}}{{dt}} = v - 1 + {t^2}\;\;\text{or}\;\; t\frac{{dv}}{{dt}} = v + {t^2} - 1.\]
We first find the solution of the corresponding homogeneous equation.
\[t\frac{{dv}}{{dt}} = v,\;\; \Rightarrow \frac{{dv}}{v} = \frac{{dt}}{t},\;\; \Rightarrow \int {\frac{{dv}}{v}} = \int {\frac{{dt}}{t}} ,\;\; \Rightarrow \ln \left| v \right| = \ln \left| t \right| + \ln C,\;\; \Rightarrow {v_0}\left( t \right) = Ct,\]
where \(C\) is an arbitrary number.
Now we determine the solution of the nonhomogeneous equation, using the method of variation of parameters:
\[v\left( t \right) = C\left( t \right)t,\;\; \Rightarrow \frac{{dv\left( t \right)}}{{dt}} = \frac{{dC\left( t \right)}}{{dt}}t + C\left( t \right).\]
After substitution we get the expression for the derivative \({\frac{{dC}}{{dt}}}:\)
\[t\left( {t\frac{{dC}}{{dt}} + C} \right) = Ct + {t^2} - 1,\;\; \Rightarrow {t^2}\frac{{dC}}{{dt}} + \cancel{Ct} = \cancel{Ct} + {t^2} - 1,\;\; \Rightarrow \frac{{dC}}{{dt}} = \frac{{{t^2} - 1}}{{{t^2}}} = 1 - \frac{1}{{{t^2}}}.\]
Integrating, we find the function \(C\left( t \right):\)
\[C\left( t \right) = \int {\left( {1 - \frac{1}{{{t^2}}}} \right)dt} = t + \frac{1}{t}.\]
Then the function \(v\left( t \right)\) will be expressed by the formula
\[v\left( t \right) = C\left( t \right)t = {t^2} + 1.\]
Now, it is easy to find the function \(u\left( t \right):\)
\[v - tu = 1,\;\; \Rightarrow tu = v - 1,\;\; \Rightarrow tu = {t^2} + \cancel{1} - \cancel{1},\;\; \Rightarrow u\left( t \right) = t.\]
So the second solution of the system is
\[{\mathbf{X}_2}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_2}\left( t \right)}\\
{{y_2}\left( t \right)}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
u\\
v
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
t\\
{{t^2} + 1}
\end{array}} \right].\]
The general solution is written as
\[\mathbf{X}\left( t \right) = {C_1}{\mathbf{X}_1}\left( t \right) + {C_2}{\mathbf{X}_2}\left( t \right)
= {C_1}\left[ {\begin{array}{*{20}{c}}
1\\
t
\end{array}} \right] + {C_2}\left[ {\begin{array}{*{20}{c}}
t\\
{{t^2} + 1}
\end{array}} \right],\]
where \({C_1},{C_2}\) are arbitrary constants.
