Differential Equations - Systems: Non-homogeneous

We now need to address non-homogeneous systems briefly. Both of the methods that we looked at back in the second order differential equations chapter can also be used here. As we will see Undetermined Coefficients is almost identical when used on systems while Variation of Parameters will need to have a new formula derived, but will actually be slightly easier when applied to systems.

Undetermined Coefficients

The method of Undetermined Coefficients for systems is pretty much identical to the second order differential equation case. The only difference is that the coefficients will need to be vectors now.

Let’s take a quick look at an example.

Example 1 Find the general solution to the following system.

{\vec{x}}^{'} = (\begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix}) \vec{x} + t (\begin{matrix} 2 \\ - 4 \end{matrix})

We already have the complementary solution as we solved that part back in the real eigenvalue section. It is,

{\vec{x}}_{c} (t) = c_{1} e^{- t} (\begin{matrix} - 1 \\ 1 \end{matrix}) + c_{2} e^{4 t} (\begin{matrix} 2 \\ 3 \end{matrix})

Guessing the form of the particular solution will work in exactly the same way it did back when we first looked at this method. We have a linear polynomial and so our guess will need to be a linear polynomial. The only difference is that the “coefficients” will need to be vectors instead of constants. The particular solution will have the form,

{\vec{x}}_{P} = t \vec{a} + \vec{b} = t (\begin{matrix} a_{1} \\ a_{2} \end{matrix}) + (\begin{matrix} b_{1} \\ b_{2} \end{matrix})

So, we need to differentiate the guess

{\vec{x}}_{P}^{'} = \vec{a} = (\begin{matrix} a_{1} \\ a_{2} \end{matrix})

Before plugging into the system let’s simplify the notation a little to help with our work. We’ll write the system as,

{\vec{x}}^{'} = (\begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix}) \vec{x} + t (\begin{matrix} 2 \\ - 4 \end{matrix}) = A \vec{x} + t \vec{g}

This will make the following work a little easier. Now, let’s plug things into the system.

\begin{aligned} \vec{a} & = A (t \vec{a} + \vec{b}) + t \vec{g} \\ \vec{a} & = t A \vec{a} + A \vec{b} + t \vec{g} \\ \vec{0} & = t (A \vec{a} + \vec{g}) + (A \vec{b} - \vec{a}) \end{aligned}

Now we need to set the coefficients equal. Doing this gives,

\begin{aligned} t^{1} : & A \vec{a} + \vec{g} & = \vec{0} & A \vec{a} & = - \vec{g} \\ t^{0} : & A \vec{b} - \vec{a} & = \vec{0} & A \vec{b} & = \vec{a} \end{aligned}

Now only

\vec{a}

is unknown in the first equation so we can use Gaussian elimination to solve the system. We’ll leave this work to you to check.

(\begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix}) (\begin{matrix} a_{1} \\ a_{2} \end{matrix}) = - (\begin{matrix} 2 \\ - 4 \end{matrix}) \Rightarrow \vec{a} = (\begin{matrix} 3 \\ - \frac{5}{2} \end{matrix})

Now that we know

\vec{a}

we can solve the second equation for

\vec{b}

(\begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix}) (\begin{matrix} b_{1} \\ b_{2} \end{matrix}) = (\begin{matrix} 3 \\ - \frac{5}{2} \end{matrix}) \Rightarrow \vec{b} = (\begin{matrix} - \frac{11}{4} \\ \frac{23}{8} \end{matrix})

So, since we were able to solve both equations, the particular solution is then,

{\vec{x}}_{P} = t (\begin{matrix} 3 \\ - \frac{5}{2} \end{matrix}) + (\begin{matrix} - \frac{11}{4} \\ \frac{23}{8} \end{matrix})

The general solution is then,

\vec{x} (t) = c_{1} e^{- t} (\begin{matrix} - 1 \\ 1 \end{matrix}) + c_{2} e^{4 t} (\begin{matrix} 2 \\ 3 \end{matrix}) + t (\begin{matrix} 3 \\ - \frac{5}{2} \end{matrix}) + (\begin{matrix} - \frac{11}{4} \\ \frac{23}{8} \end{matrix})

So, as you can see undetermined coefficients is nearly the same as the first time we saw it. The work in solving for the “constants” is a little messier however.

Variation of Parameters

In this case we will need to derive a new formula for variation of parameters for systems. The derivation this time will be much simpler than the when we first saw variation of parameters.

First let

X (t)

be a matrix whose i^th column is the i^th linearly independent solution to the system,

{\vec{x}}^{'} = A \vec{x}

Now it can be shown that

X (t)

will be a solution to the following differential equation.

\begin{matrix}  \end{matrix}

This is nothing more than the original system with the matrix in place of the original vector.

We are going to try and find a particular solution to

{\vec{x}}^{'} = A \vec{x} + \vec{g} (t)

We will assume that we can find a solution of the form,

{\vec{x}}_{P} = X (t) \vec{v} (t)

where we will need to determine the vector

\vec{v} (t)

. To do this we will need to plug this into the non-homogeneous system. Don’t forget to product rule the particular solution when plugging the guess into the system.

X^{'} \vec{v} + X {\vec{v}}^{'} = A X \vec{v} + \vec{g}

Note that we dropped the

(t)

part of things to simplify the notation a little. Now using

(1)

we can rewrite this a little.

\begin{aligned} X^{'} \vec{v} + X {\vec{v}}^{'} & = X^{'} \vec{v} + \vec{g} \\ X {\vec{v}}^{'} & = \vec{g} \end{aligned}

Because we formed

X

using linearly independent solutions we know that

det (X)

must be nonzero and this in turn means that we can find the inverse of

X

. So, multiply both sides by the inverse of

X

{\vec{v}}^{'} = X^{- 1} \vec{g}

Now all that we need to do is integrate both sides to get

\vec{v} (t)

\vec{v} (t) = \int X^{- 1} \vec{g} d t

As with the second order differential equation case we can ignore any constants of integration. The particular solution is then,

\begin{matrix}  \end{matrix}

Let’s work a quick example using this.

Example 2 Find the general solution to the following system.

{\vec{x}}^{'} = (\begin{matrix} - 5 & 1 \\ 4 & - 2 \end{matrix}) \vec{x} + e^{2 t} (\begin{matrix} 6 \\ - 1 \end{matrix})

We found the complementary solution to this system in the real eigenvalue section. It is,

{\vec{x}}_{c} (t) = c_{1} e^{- t} (\begin{matrix} 1 \\ 4 \end{matrix}) + c_{2} e^{- 6 t} (\begin{matrix} - 1 \\ 1 \end{matrix})

Now the matrix

X

is,

X = (\begin{matrix} e^{- t} & - e^{- 6 t} \\ 4 e^{- t} & e^{- 6 t} \end{matrix})

Now, we need to find the inverse of this matrix. We saw how to find inverses of matrices back in the second linear algebra review section and the process is the same here even though we don’t have constant entries. We’ll leave the detail to you to check.

X^{- 1} = (\begin{matrix} \frac{1}{5} e^{t} & \frac{1}{5} e^{t} \\ - \frac{4}{5} e^{6 t} & \frac{1}{5} e^{6 t} \end{matrix})

Now do the multiplication in the integral.

X^{- 1} \vec{g} = (\begin{matrix} \frac{1}{5} e^{t} & \frac{1}{5} e^{t} \\ - \frac{4}{5} e^{6 t} & \frac{1}{5} e^{6 t} \end{matrix}) (\begin{matrix} 6 e^{2 t} \\ - e^{2 t} \end{matrix}) = (\begin{matrix} e^{3 t} \\ - 5 e^{8 t} \end{matrix})

Now do the integral.

\int X^{- 1} \vec{g} d t = \int (\begin{matrix} e^{3 t} \\ - 5 e^{8 t} \end{matrix}) d t = (\begin{matrix} \int e^{3 t} d t \\ \int - 5 e^{8 t} d t \end{matrix}) = (\begin{matrix} \frac{1}{3} e^{3 t} \\ - \frac{5}{8} e^{8 t} \end{matrix})

Remember that to integrate a matrix or vector you just integrate the individual entries.
We can now get the particular solution.

\begin{aligned} {\vec{x}}_{P} & = X \int X^{- 1} \vec{g} d t \\ = (\begin{matrix} e^{- t} & - e^{- 6 t} \\ 4 e^{- t} & e^{- 6 t} \end{matrix}) (\begin{matrix} \frac{1}{3} e^{3 t} \\ - \frac{5}{8} e^{8 t} \end{matrix}) \\ = (\begin{matrix} \frac{23}{24} e^{2 t} \\ \frac{17}{24} e^{2 t} \end{matrix}) \\ = e^{2 t} (\begin{matrix} \frac{23}{24} \\ \frac{17}{24} \end{matrix}) \end{aligned}

The general solution is then,

\vec{x} (t) = c_{1} e^{- t} (\begin{matrix} 1 \\ 4 \end{matrix}) + c_{2} e^{- 6 t} (\begin{matrix} - 1 \\ 1 \end{matrix}) + e^{2 t} (\begin{matrix} \frac{23}{24} \\ \frac{17}{24} \end{matrix})

So, some of the work can be a little messy, but overall not too bad.

We looked at two methods of solving non-homogeneous differential equations here and while the work can be a little messy they aren’t too bad. Of course, we also kept the non-homogeneous part fairly simple here. More complicated problems will have significant amounts of work involved.

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