# 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.