6.8: Nonhomogeneous Systems

Before leaving the theory of systems of linear, constant coefficient systems, we will discuss nonhomogeneous systems. We would like to solve systems of the form

\[\mathbf{x}^{\prime}=A(t) \mathbf{x}+\mathbf{f}(t) \nonumber \]

We will assume that we have found the fundamental matrix solution of the homogeneous equation. Furthermore, we will assume that \(A(t)\) and \(\mathbf{f}(t)\) are continuous on some common domain.

As with second order equations, we can look for solutions that are a sum of the general solution to the homogeneous problem plus a particular solution of the nonhomogeneous problem. Namely, we can write the general solution as

\[\mathbf{x}(t)=\Phi(t) \mathbf{C}+\mathbf{x}_{p}(t)\nonumber \]

where \(C\) is an arbitrary constant vector, \(\Phi(t)\) is the fundamental matrix solution of \(\mathbf{x}^{\prime}=A(t) \mathbf{x}\), and

\[\mathbf{x}_{p}^{\prime}=A(t) \mathbf{x}_{p}+\mathbf{f}(t)\nonumber \]

Such a representation is easily verified.

We need to find the particular solution, \(\mathbf{x}_{p}(t)\). We can do this by applying The Method of Variation of Parameters for Systems. We consider a solution in the form of the solution of the homogeneous problem, but replace the constant vector by unknown parameter functions. Namely, we assume that

\[\mathbf{x}_{p}(t)=\Phi(t) \mathbf{c}(t)\nonumber \]

Differentiating, we have that

\[\mathbf{x}_{p}^{\prime}=\Phi_{\mathbf{c}}^{\prime} \mathbf{c} \Phi \mathbf{c}^{\prime}=A \Phi \mathbf{c}+\Phi \mathbf{c}^{\prime} \nonumber \]

Or

\[\mathbf{x}_{p}^{\prime}-A \mathbf{x}_{p}=\Phi \mathbf{c}^{\prime}\nonumber \]

But the left side is \(\mathbf{f . ~ S o , ~ w e ~ h a v e ~ t h a t , ~}\)

\[\Phi \mathbf{c}^{\prime}=\mathbf{f}\nonumber \]

or, since \(\Phi\) is invertible (why?),

\[\mathbf{c}^{\prime}=\Phi^{-1} \mathbf{f}\nonumber \]

In principle, this can be integrated to give c. Therefore, the particular solution can be written as

\[\mathbf{x}_{p}(t)=\Phi(t) \int^{t} \Phi^{-1}(s) \mathbf{f}(s) d s \nonumber \]

This is the variation of parameters formula.

The general solution of Equation \(\PageIndex{2}\) has been found as

\[\mathbf{x}(t)=\Phi(t) \mathbf{C}+\Phi(t) \int^{t} \Phi^{-1}(s) \mathbf{f}(s) d s \nonumber \]

We can use the general solution to find the particular solution of an initial value problem consisting of Equation \(\PageIndex{2}\) and the initial condition \(\mathbf{x}\left(t_{0}\right)=\mathbf{x}_{0} .\) This condition is satisfied for a solution of the form

\[\mathbf{x}(t)=\Phi(t) \mathbf{C}+\Phi(t) \int_{t_{0}}^{t} \Phi^{-1}(s) \mathbf{f}(s) d s \nonumber \]

provided

\[\mathbf{x}_{0}=\mathbf{x}\left(t_{0}\right)=\Phi\left(t_{0}\right) \mathbf{C}. \nonumber \]

This can be solved for \(\mathrm{C}\) as in the last section. Inserting the solution back into the general solution Equation \(\PageIndex{4}\), we have

\[\mathbf{x}(t)=\Phi(t) \Phi^{-1}\left(t_{0}\right) \mathbf{x}_{0}+\Phi(t) \int_{t_{0}}^{t} \Phi^{-1}(s) \mathbf{f}(s) d s \nonumber \]

This solution can be written a little neater in terms of the principal matrix solution, \(\Psi(t)=\Phi(t) \Phi^{-1}\left(t_{0}\right)\) :

\[\mathbf{x}(t)=\Psi(t) \mathbf{x}_{0}+\Psi(t) \int_{t_{0}}^{t} \Psi^{-1}(s) \mathbf{f}(s) d s \nonumber \]

Finally, one further simplification occurs when \(A\) is a constant matrix, which are the only types of problems we have solved in this chapter. In this case, we have that \(\Psi^{-1}(t)=\Psi(-t) .\) So, computing \(\Psi^{-1}(t)\) is relatively easy.

Example \(\PageIndex{1}\)

\(x^{\prime \prime}+x=2 \cos t, x(0)=4, x^{\prime}(0)=0 .\) This example can be solved using the Method of Undetermined Coefficients. However, we will use the matrix method described in this section.

First, we write the problem in matrix form. The system can be Written as

\[ \begin{gathered} x^{\prime}=y \\ y^{\prime}=-x+2 \cos t \end{gathered} \label{6.106} \]

Thus, we have a nonhomogeneous system of the form

\[\mathbf{x}^{\prime}=A \mathbf{x}+\mathbf{f}=\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)\left(\begin{array}{l} x \\ y \end{array}\right)+\left(\begin{array}{c} 0 \\ 2 \cos t \end{array}\right) \nonumber \]

Next we need the fundamental matrix of solutions of the homogeneous problem. We have that

\[A=\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right) \nonumber \]

The eigenvalues of this matrix are \(\lambda=\pm i .\) An eigenvector associated with \(\lambda=i\) is easily found as \(\left(\begin{array}{c}1 \\ i\end{array}\right)\). This leads to a complex solution

\[\left(\begin{array}{l} 1 \\ i \end{array}\right) e^{i t}=\left(\begin{array}{c} \cos t+i \sin t \\ i \cos t-\sin t \end{array}\right)\nonumber \]

From this solution we can construct the fundamental solution matrix

\[\Phi(t)=\left(\begin{array}{cc} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right)\nonumber \]

So, the general solution to the homogeneous problem is

\[\mathbf{x}_{h}=\Phi(t) \mathbf{C}=\left(\begin{array}{c} c_{1} \cos t+c_{2} \sin t \\ -c_{1} \sin t+c_{2} \cos t \end{array}\right)\nonumber \]

Next we seek a particular solution to the nonhomogeneous problem. From Equation \(\PageIndex{4}\) we see that we need \(\Phi^{-1}(s) \mathbf{f}(s)\). Thus, we have

\[ \begin{aligned} \Phi^{-1}(s) \mathbf{f}(s) &=\left(\begin{array}{cc} \cos s & -\sin s \\ \sin s & \cos s \end{array}\right)\left(\begin{array}{c} 0 \\ 2 \cos s \end{array}\right) \\ &=\left(\begin{array}{c} -2 \sin s \cos s \\ 2 \cos ^{2} s \end{array}\right) \end{aligned} \label{6.107} \]

We now compute

\[ \begin{aligned} \Phi(t) \int_{t_{0}}^{t} \Phi^{-1}(s) \mathbf{f}(s) d s &=\left(\begin{array}{cc} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right) \int_{t_{0}}^{t}\left(\begin{array}{c} -2 \sin s \cos s \\ 2 \cos ^{2} s \end{array}\right) d s \\ &=\left(\begin{array}{cc} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right)\left(\begin{array}{c} -\sin ^{2} t \\ t+\dfrac{1}{2} \sin (2 t) \end{array}\right) \\ &=\left(\begin{array}{c} t \sin t \\ \sin t+t \cos t \end{array}\right) \end{aligned} \label{6.108} \]

therefore, the general solution is

\[\mathbf{x}=\left(\begin{array}{c} c_{1} \cos t+c_{2} \sin t \\ -c_{1} \sin t+c_{2} \cos t \end{array}\right)+\left(\begin{array}{c} t \sin t \\ \sin t+t \cos t \end{array}\right)\nonumber \]

The solution to the initial value problem is

\[\mathbf{x}=\left(\begin{array}{cc} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right)\left(\begin{array}{c} 4 \\ 0 \end{array}\right)+\left(\begin{array}{c} t \sin t \\ \sin t+t \cos t \end{array}\right)\nonumber \]

Or

\[\mathbf{x}=\left(\begin{array}{c} 4 \cos t+t \sin t \\ -3 \sin t+t \cos t \end{array}\right)\nonumber \]