Differential equations – Systems of linear differential equations – Nonhomogeneous systems

Now we consider nonhomogeneous systems of the form

\[\mathbf{x}'(t)=A\mathbf{x}(t)+\mathbf{g}(t)\quad\text{with}\quad\mathbf{g}(t)\neq\mathbf{0}.\]

The general solution has the form \(\mathbf{x}(t)=\mathbf{x}_p(t)+\mathbf{x}_h(t)\) with \(\mathbf{x}_h(t)\) the general solution of the homogeneous sytem \(\mathbf{x}'(t)=A\mathbf{x}(t)\) and \(\mathbf{x}_p(t)\) a particular solution of the nonhomogeneous system. We consider three different methods:

1. Uncoupling; in principle this is only possible if the matrix \(A\) is diagonalizable.


2. The method of undetermined coefficients.


3. The method of variation of parameters.

Uncoupling

If the matrix \(A\) is diagonizable, then we have: \(A=PDP^{-1}\) for some invertible matrix \(P\) and diagonal matrix \(D\). Now let \(\mathbf{x}=P\mathbf{y}(t)\), then we have:

\[\mathbf{x}'(t)=A\mathbf{x}(t)+\mathbf{g}(t)\quad\Longleftrightarrow\quad\mathbf{y}'(t)=D\mathbf{y}(t)+\mathbf{h}(t) \quad\text{met}\quad\mathbf{h}(t)=P^{-1}\mathbf{g}(t).\]

The system for \(\mathbf{y}(t)\) is then an uncoupled system: a system of noncoupled (first order linear) differential equations. We solve this system for every differential equation separately. Thereafter we then obtain: \(\mathbf{x}=P^{-1}\mathbf{y}(t)\).

Defect matrices

If the matrix \(A\) is defect, then \(A\) is not diagonizable. However, we have: \(A=PJP^{-1}\) for some invertible matrix \(P\) and Jordan normal form \(J\). Now let \(\mathbf{x}=P\mathbf{y}(t)\), then we have:

\[\mathbf{x}'(t)=A\mathbf{x}(t)+\mathbf{g}(t)\quad\Longleftrightarrow\quad\mathbf{y}'(t)=J\mathbf{y}(t)+\mathbf{h}(t) \quad\text{met}\quad\mathbf{h}(t)=P^{-1}\mathbf{g}(t).\]

Although this system is not (completely) uncoupled, it is often much easier than the original system.

Method of undetermined coefficients

First find the general solution \(\mathbf{x}_h(t)\) of the homogeneous system \(\mathbf{x}'(t)=A\mathbf{x}(t)\). Thereafter, based on \(\mathbf{g}(t)\) and the general solution \(\mathbf{x}_h(t)\) of the homogeneous system \(\mathbf{x}'(t)=A\mathbf{x}(t)\), we can sometimes determine an appropriate form of a particular solution \(\mathbf{x}_p(t)\) of the nonhomogeneous system. In this form there appear coefficients which should be determined. This only works if \(\mathbf{g}(t)\) consists of exponential functions, polynomials, sines and cosines, and combinations of these functions.

Method of variation of parameters

Find the general solution \(\mathbf{x}(t)=\Psi(t)\mathbf{c}\) of the homogeneous system \(\mathbf{x}'(t)=A\mathbf{x}(t)\). So hereby \(\Psi(t)\) is a fundamental matrix of the homogeneous system \(\mathbf{x}'(t)=A\mathbf{x}(t)\). Now let \(\mathbf{x}(t)=\Psi(t)\mathbf{u}(t)\), then we have: \(\mathbf{x}'(t)=\Psi'(t)\mathbf{u}(t)+\Psi(t)\mathbf{u}'(t)\). Substitution then gives

\[\Psi'(t)\mathbf{u}(t)+\Psi(t)\mathbf{u}'(t)=A\Psi(t)\mathbf{u}(t)+\mathbf{g}(t)\quad\Longrightarrow\quad\Psi(t)\mathbf{u}'(t)=\mathbf{g}(t).\]

After all: \(\Psi'(t)=A\Psi(t)\). Since \(\Psi(t)\) is a fundamental matrix, it is invertible. Hence: \(\mathbf{u}'(t)=\Psi^{-1}(t)\mathbf{g}(t)\).

This implies: \(\mathbf{u}(t)=\displaystyle\int\Psi^{-1}(t)\mathbf{g}(t)\,dt\). This is not always an easy step. Depending on \(\mathbf{g}(t)\) and \(\Psi(t)\) this might be a rather nasty integral. However, if \(\mathbf{u}(t)\) can yet be obtained, then we finally have that \(\mathbf{x}(t)=\Psi(t)\mathbf{u}(t)\).

Example:

Consider \(\mathbf{x}'(t)=A\mathbf{x}(t)+\mathbf{g}(t)\) with \(A=\begin{pmatrix}5&-3\\6&-4\end{pmatrix}\) and \(\mathbf{g}(t)=\begin{pmatrix}e^{-t}\\e^t+2t+3\end{pmatrix}\). Then we have:

\[|A-rI|=\begin{vmatrix}5-r&-3\\6&-4-r\end{vmatrix}=r^2-r-2=(r-2)(r+1).\]

Hence: \(r_1=2\) and \(r_2=-1\) are the eigenvalues of \(A\). Further we have:

\[r_1=2:\quad\begin{pmatrix}3&-3\\6&-6\end{pmatrix}\sim\begin{pmatrix}1&-1\\0&0\end{pmatrix}\quad\Longrightarrow\quad\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}\]

and

\[r_2=-1:\quad\begin{pmatrix}6&-3\\6&-3\end{pmatrix}\sim\begin{pmatrix}2&-1\\0&0\end{pmatrix}\quad\Longrightarrow\quad\mathbf{v}_2=\begin{pmatrix}1\\2\end{pmatrix}.\]

Hence: \(\mathbf{x}_h(t)=c_1\mathbf{v}_1e^{r_1t}+c_2\mathbf{v}_2e^{r_2t}=c_1\begin{pmatrix}1\\1\end{pmatrix}e^{2t}+c_2\begin{pmatrix}1\\2\end{pmatrix}e^{-t}\) with \(c_1,c_2\in\mathbb{R}\).

This can also be written as \(\mathbf{x}_h(t)=\Psi(t)\mathbf{c}\) with \(\mathbf{c}=\begin{pmatrix}c_1\\c_2\end{pmatrix}\) and \(\Psi(t)=\begin{pmatrix}e^{2t}&e^{-t}\\e^{2t}&2e^{-t}\end{pmatrix}\), a fundamental matrix of \(\mathbf{x}'(t)=A\mathbf{x}(t)\).

The matrix \(A\) is diagonalizable: \(A=PDP^{-1}\) with \(P=\begin{pmatrix}1&1\\1&2\end{pmatrix}\) and \(D=\begin{pmatrix}2&0\\0&-1\end{pmatrix}\). Then we have: \(P^{-1}=\begin{pmatrix}2&-1\\-1&1\end{pmatrix}\).

1) Let \(\mathbf{x}(t)=P\mathbf{y}(t)\), then we have: \(\mathbf{y}'(t)=D\mathbf{y}(t)+\mathbf{h}(t)\) with \(\mathbf{h}(t)=P^{-1}\mathbf{g}(t) =\begin{pmatrix}2e^{-t}-e^t-2t-3\\-e^{-t}+e^t+2t+3\end{pmatrix}\). This is a system of noncoupled differential equations:

\[\left\{\begin{array}{l}y_1'(t)=2y_1(t)+2e^{-t}-e^t-2t-3\\[2.5mm]y_2'(t)=-y_2(t)-e^{-t}+e^t+2t+3\end{array}\right.\]

These can be solved using the theory of first-order linear differential equations, however we can also apply the method of undetermined coefficients:

\[y_1'(t)=2y_1(t)+2e^{-t}-e^t-2t-3\quad\Longrightarrow\quad y_{1,h}(t)=c_1e^{2t}\quad\text{and}\quad y_{1,p}(t)=Ae^{-t}+Be^t+Ct+D.\]

Substitution then gives: \(-Ae^{-t}+Be^t+C=2Ae^{t}+2Be^t+2Ct+2D+2e^{-t}-e^t-2t-3\). This implies: \(A=-\frac{2}{3}\), \(B=C=1\) and \(D=2\). Hence: \(y_{1,p}(t)=-\frac{2}{3}e^{-t}+e^t+t+2\).

\[y_2'(t)=-y_2(t)-e^{-t}+e^t+2t+3\quad\Longrightarrow\quad y_{2,h}(t)=c_2e^{-t}\quad\text{and}\quad y_{2,p}(t)=Ate^{-t}+Be^t+Ct+D.\]

Substitution then gives: \(A(1-t)e^{-t}+Be^t+C=-Ate^{-t}-Be^t-Ct-D-e^{-t}+e^t+2t+3\). This implies: \(A=-1\), \(B=\frac{1}{2}\), \(C=2\) and \(D=1\). Hence: \(y_{2,p}(t)=-te^{-t}+\frac{1}{2}e^t+2t+1\). So we have obtained that

\[\left\{\begin{array}{l}y_1(t)=-\frac{2}{3}e^{-t}+e^t+t+2+c_1e^{2t}\\[0.5mm]y_2(t)=-te^{-t}+\frac{1}{2}e^t+2t+1+c_2e^{-t}\end{array}\right. \quad\Longleftrightarrow\quad\mathbf{y}(t)=\begin{pmatrix}y_1(t)\\y_2(t)\end{pmatrix}=\begin{pmatrix}-\frac{2}{3}e^{-t}+e^t+t+2+c_1e^{2t}\\ -te^{-t}+\frac{1}{2}e^t+2t+1+c_2e^{-t}\end{pmatrix}.\]

Finally we have:

\begin{align*} \mathbf{x}(t)=P\mathbf{y}(t)&=\begin{pmatrix}1&1\\1&2\end{pmatrix}\begin{pmatrix}-\frac{2}{3}e^{-t}+e^t+t+2+c_1e^{2t}\\ -te^{-t}+\frac{1}{2}e^t+2t+1+c_2e^{-t}\end{pmatrix}=\begin{pmatrix}-te^{-t}-\frac{2}{3}e^{-t}+\frac{3}{2}e^t+3t+3+c_1e^{2t}+c_2e^{-t}\\ -2te^{-t}-\frac{2}{3}e^{-t}+2e^t+5t+4+c_1e^{2t}+2c_2e^{-t}\end{pmatrix}\\[2.5mm] &=-\begin{pmatrix}1\\2\end{pmatrix}te^{-t}-\frac{2}{3}\begin{pmatrix}1\\1\end{pmatrix}e^{-t}+\frac{1}{2}\begin{pmatrix}3\\4\end{pmatrix}e^t +\begin{pmatrix}3\\5\end{pmatrix}t+\begin{pmatrix}3\\4\end{pmatrix}+c_1\begin{pmatrix}1\\1\end{pmatrix}e^{2t}+c_2\begin{pmatrix}1\\2\end{pmatrix}e^{-t}. \end{align*}

2) Based on \(\mathbf{x}_h(t)=c_1\begin{pmatrix}1\\1\end{pmatrix}e^{2t}+c_2\begin{pmatrix}1\\2\end{pmatrix}e^{-t}\) and \(\mathbf{g}(t)=\begin{pmatrix}e^{-t}\\e^t+2t+3\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+\begin{pmatrix}0\\1\end{pmatrix}e^t +\begin{pmatrix}0\\2\end{pmatrix}t+\begin{pmatrix}0\\3\end{pmatrix}\) we choose:

\[\mathbf{x}_p(t)=\mathbf{v}_1te^{-t}+\mathbf{v}_2e^{-t}+\mathbf{v}_3e^t+\mathbf{v}_4t+\mathbf{v}_5.\]

Substitution then gives:

\[\mathbf{v}_1(1-t)e^{-t}-\mathbf{v}_2e^{-t}+\mathbf{v}_3e^t+\mathbf{v}_4=A\mathbf{v}_1te^{-t}+A\mathbf{v}_2e^{-t}+A\mathbf{v}_3e^t+A\mathbf{v}_4t+A\mathbf{v}_5 +\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+\begin{pmatrix}0\\1\end{pmatrix}e^t+\begin{pmatrix}0\\2\end{pmatrix}t+\begin{pmatrix}0\\3\end{pmatrix}.\]

This implies:

\[A\mathbf{v}_1=-\mathbf{v}_1,\quad A\mathbf{v}_2+\begin{pmatrix}1\\0\end{pmatrix}=\mathbf{v}_1-\mathbf{v}_2,\quad A\mathbf{v}_3+\begin{pmatrix}0\\1\end{pmatrix}=\mathbf{v}_3,\quad A\mathbf{v}_4+\begin{pmatrix}0\\2\end{pmatrix}=\mathbf{0} \quad\text{and}\quad A\mathbf{v}_5+\begin{pmatrix}0\\3\end{pmatrix}=\mathbf{v}_4.\]

Or equivalently:

\[(A+I)\mathbf{v}_1=\mathbf{0},\quad(A+I)\mathbf{v}_2=\mathbf{v}_1-\begin{pmatrix}1\\0\end{pmatrix},\quad (A-I)\mathbf{v}_3=-\begin{pmatrix}0\\1\end{pmatrix},\quad A\mathbf{v}_4=-\begin{pmatrix}0\\2\end{pmatrix}\quad\text{and}\quad A\mathbf{v}_5=\mathbf{v}_4-\begin{pmatrix}0\\3\end{pmatrix}.\]

Hence: \(\mathbf{v}_1\) is an eigenvector of \(A\) corresponding to the eigenvalue \(-1\): \(\mathbf{v}_1=\alpha\begin{pmatrix}1\\2\end{pmatrix}\). Then we have:

\[(A+I)\mathbf{v}_2=\mathbf{v}_1-\begin{pmatrix}1\\0\end{pmatrix}:\quad\left(\left.\begin{matrix}6&-3\\6&-3\end{matrix}\,\right|\,\begin{matrix}\alpha-1\\2\alpha\end{matrix}\right) \sim\left(\left.\begin{matrix}2&-1\\0&0\end{matrix}\,\right|\,\begin{matrix}\frac{2}{3}\alpha\\\alpha+1\end{matrix}\right) \quad\Longrightarrow\quad\alpha=-1\quad\text{en}\quad\mathbf{v}_2=-\frac{2}{3}\begin{pmatrix}1\\1\end{pmatrix}\quad\text{bijvoorbeeld.}\]

Further we have:

\[(A-I)\mathbf{v}_3=-\begin{pmatrix}0\\1\end{pmatrix}:\quad\left(\left.\begin{matrix}4&-3\\6&-5\end{matrix}\,\right|\,\begin{matrix}0\\-1\end{matrix}\right) \quad\Longrightarrow\quad\mathbf{v}_3=\frac{1}{2}\begin{pmatrix}3\\4\end{pmatrix},\] \[A\mathbf{v}_4=-\begin{pmatrix}0\\2\end{pmatrix}:\quad\left(\left.\begin{matrix}5&-3\\6&-4\end{matrix}\,\right|\,\begin{matrix}0\\-2\end{matrix}\right) \quad\Longrightarrow\quad\mathbf{v}_4=\begin{pmatrix}3\\5\end{pmatrix}\]

and

\[A\mathbf{v}_5=\mathbf{v}_4-\begin{pmatrix}0\\3\end{pmatrix}:\left(\left.\begin{matrix}5&-3\\6&-4\end{matrix}\,\right|\,\begin{matrix}3\\2\end{matrix}\right) \quad\Longrightarrow\quad\mathbf{v}_5=\begin{pmatrix}3\\4\end{pmatrix}.\]

Hence:

\[\mathbf{x}(t)=-\begin{pmatrix}1\\2\end{pmatrix}te^{-t}-\frac{2}{3}\begin{pmatrix}1\\1\end{pmatrix}e^{-t}+\frac{1}{2}\begin{pmatrix}3\\4\end{pmatrix}e^t +\begin{pmatrix}3\\5\end{pmatrix}t+\begin{pmatrix}3\\4\end{pmatrix}+c_1\begin{pmatrix}1\\1\end{pmatrix}e^{2t}+c_2\begin{pmatrix}1\\2\end{pmatrix}e^{-t}.\]

3) We have: \(\mathbf{x}_h(t)=\Psi(t)\mathbf{c}\) with \(\mathbf{c}=\begin{pmatrix}c_1\\c_2\end{pmatrix}\) and \(\Psi(t)=\begin{pmatrix}e^{2t}&e^{-t}\\e^{2t}&2e^{-t}\end{pmatrix}\). Then we have:

\[\Psi(t)=\begin{pmatrix}e^{2t}&e^{-t}\\e^{2t}&2e^{-t}\end{pmatrix}=\begin{pmatrix}1&1\\1&2\end{pmatrix}\begin{pmatrix}e^{2t}&0\\0&e^{-t}\end{pmatrix}\quad\Longrightarrow\quad \Psi^{-1}(t)=\begin{pmatrix}e^{-2t}&0\\0&e^t\end{pmatrix}\begin{pmatrix}2&-1\\-1&1\end{pmatrix}=\begin{pmatrix}2e^{-2t}&-e^{-2t}\\-e^t&e^t\end{pmatrix}.\]

Now let: \(\mathbf{x}(t)=\Psi(t)\mathbf{u}(t)\). Then we have: \(\mathbf{x}'(t)=\Psi'(t)\mathbf{u}(t)+\Psi(t)\mathbf{u}'(t)\). Substitution then gives:

\[\Psi'(t)\mathbf{u}(t)+\Psi(t)\mathbf{u}'(t)=A\Psi(t)\mathbf{u}(t)+\mathbf{g}(t)\quad\Longleftrightarrow\quad\Psi(t)\mathbf{u}'(t)=\mathbf{g}(t) \quad\Longleftrightarrow\quad\mathbf{u}'(t)=\Psi^{-1}(t)\mathbf{g}(t).\]

After all: \(\Psi'(t)=A\Psi(t)\). Hence:

\[\mathbf{u}'(t)=\Psi^{-1}(t)\mathbf{g}(t)=\begin{pmatrix}2e^{-2t}&-e^{-2t}\\-e^t&e^t\end{pmatrix}\begin{pmatrix}e^{-t}\\e^t+2t+3\end{pmatrix} =\begin{pmatrix}2e^{-3t}-e^{-t}-(2t+3)e^{-2t}\\-1+e^{2t}+(2t+3)e^t\end{pmatrix}.\]

Using integration by parts we obtain

\[\int(2t+3)e^{-2t}\,dt=-\frac{1}{2}\int(2t+3)\,de^{-2t}=-\frac{1}{2}(2t+3)e^{-2t}+\int e^{-2t}\,dt=-te^{-2t}-\frac{3}{2}e^{-2}-\frac{1}{2}e^{-2t}+C =-(t+2)e^{-2t}+C\]

and

\[\int(2t+3)e^t\,dt=\int(2t+3)\,de^t=(2t+3)e^t-2\int e^t\,dt=2te^t+3e^t-2e^t+K=(2t+1)e^t+K.\]

Hence:

\[\mathbf{u}(t)=\int\begin{pmatrix}2e^{-3t}-e^{-t}-(2t+3)e^{-2t}\\-1+e^{2t}+(2t+3)e^t\end{pmatrix}\,dt =\begin{pmatrix}-\frac{2}{3}e^{-3t}+e^{-t}+(t+2)e^{-2t}+c_1\\-t+\frac{1}{2}e^{2t}+(2t+1)e^t+c_2\end{pmatrix}.\]

Finally we have:

\begin{align*} \mathbf{x}(t)=\Psi(t)\mathbf{u}(t)&=\begin{pmatrix}e^{2t}&e^{-t}\\e^{2t}&2e^{-t}\end{pmatrix} \begin{pmatrix}-\frac{2}{3}e^{-3t}+e^{-t}+(t+2)e^{-2t}+c_1\\-t+\frac{1}{2}e^{2t}+(2t+1)e^t+c_2\end{pmatrix} =\begin{pmatrix}-te^{-t}-\frac{2}{3}e^{-t}+\frac{3}{2}e^t+3t+3+c_1e^{2t}+c_2e^{-t}\\ -2te^{-t}-\frac{2}{3}e^{-t}+2e^t+5t+4+c_1e^{2t}+2c_2e^{-t}\end{pmatrix}\\[2.5mm] &=-\begin{pmatrix}1\\2\end{pmatrix}te^{-t}-\frac{2}{3}\begin{pmatrix}1\\1\end{pmatrix}e^{-t}+\frac{1}{2}\begin{pmatrix}3\\4\end{pmatrix}e^t +\begin{pmatrix}3\\5\end{pmatrix}t+\begin{pmatrix}3\\4\end{pmatrix}+c_1\begin{pmatrix}1\\1\end{pmatrix}e^{2t}+c_2\begin{pmatrix}1\\2\end{pmatrix}e^{-t}. \end{align*}

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Last modified on July 1, 2021