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8.3 Nonhomogeneous Linear Systems: Class Notes

8.3a

$𝐗\prime =𝐀𝐗+𝐅(t)$ is a nonhomogeneous system if $𝐅(t)\ne \mathbf{0}$.

Example 1

Use the method of undetermined coefficients to solve the given nonhomogeneous system.

$$𝐗\prime =\left(\begin{array}{cc}1& -4\\ 4& 1\end{array}\right)𝐗+\left(\begin{array}{c}4t+9e^{6t}\\ -t+e^{6t}\end{array}\right)$$

Solution

Step 1: Solve $𝐗\prime =𝐀𝐗$.

$$\left|\begin{array}{cc}1-\lambda & -4\\ 4& 1-\lambda \end{array}\right|=0⟹(\lambda -1{)}^2=-16⟹\lambda =1\pm 4i$$

For $\lambda =1+4i$:

$$\left(\begin{array}{cc}-4i& -4\\ 4& -4i\end{array}\right)𝐊=\mathbf{0}⟹k_1=i{k}_2$$

Let $k_2=-i$. Then $k_1=1$, so

$$𝐊=\left(\begin{array}{c}1\\ -i\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ -1\end{array}\right)i$$

By the theorem from 8.2 (with $\alpha =1,\beta =4$):

$${𝐗}_c=c_1{e}^t[\left(\begin{array}{c}1\\ 0\end{array}\right)\cos 4t-\left(\begin{array}{c}0\\ -1\end{array}\right)\sin 4t]+c_2{e}^t[\left(\begin{array}{c}0\\ -1\end{array}\right)\cos 4t+\left(\begin{array}{c}1\\ 0\end{array}\right)\sin 4t]$$

Step 2: Guess the form of ${𝐗}_p$. Split the forcing term:

$$𝐅(t)=\left(\begin{array}{c}4\\ -1\end{array}\right)t+\left(\begin{array}{c}9\\ 1\end{array}\right)e^{6t}$$

This is linear plus exponential, like $At+B$ plus $C{e}^{6t}$ for a single equation, so guess vectors of unknown coefficients in the same pattern:

$${𝐗}_p=\left(\begin{array}{c}{a}_3\\ b_3\end{array}\right)t+\left(\begin{array}{c}{a}_2\\ b_2\end{array}\right)+\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)e^{6t}$$

Step 3: Substitute into $𝐗\prime =𝐀𝐗+𝐅$. First the right side; multiply each row of $𝐀$ by ${𝐗}_p$ written as one column:

$$\begin{array}{cc}\hfill {\displaystyle 𝐀{𝐗}_p}& {\displaystyle =\left(\begin{array}{cc}1& -4\\ 4& 1\end{array}\right)\left(\begin{array}{c}{a}_{3}t+a_2+a_1{e}^{6t}\\ b_{3}t+b_2+b_1{e}^{6t}\end{array}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}{a}_{3}t+a_2+a_1{e}^{6t}-4b_{3}t-4b_2-4b_1{e}^{6t}\\ 4a_{3}t+4a_2+4a_1{e}^{6t}+b_{3}t+b_2+b_1{e}^{6t}\end{array}\right)}\hfill \end{array}$$

Group the $t$ terms, constant terms, and exponential terms. The left side is ${𝐗}_p^{\prime }=\left(\begin{array}{c}{a}_3\\ b_3\end{array}\right)+\left(\begin{array}{c}6a_1\\ 6b_1\end{array}\right)e^{6t}$, so

$$\begin{array}{cc}\hfill {\displaystyle \left(\begin{array}{c}{a}_3\\ b_3\end{array}\right)+\left(\begin{array}{c}6a_1\\ 6b_1\end{array}\right)e^{6t}}& {\displaystyle =\left(\begin{array}{c}{a}_3-4b_3\\ 4a_3+b_3\end{array}\right)t+\left(\begin{array}{c}{a}_2-4b_2\\ 4a_2+b_2\end{array}\right)+\left(\begin{array}{c}{a}_1-4b_1\\ 4a_1+b_1\end{array}\right)e^{6t}}\hfill \\ \hfill {\displaystyle }& {\displaystyle +\left(\begin{array}{c}4t+9e^{6t}\\ -t+e^{6t}\end{array}\right)}\hfill \end{array}$$

Match coefficients on each side:

In the constant row, use $a_3=0$ and $b_3=1$ from the $t$ row. Write ${𝐗}_c$ as a matrix times $\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)$ and add ${𝐗}_p$:

8.3b

Variation of parameters

The complementary solution is ${𝐗}_c(t)=𝚽(t)𝐂$. ($𝚽$ is the Greek letter “phi.”)

Definition: $𝚽(t)$ is the fundamental matrix of the system.

Its determinant, $det(𝚽)$, is the Wronskian.

For a system with independent solutions, the Wronskian is nonzero, so ${𝚽}^{-1}$ exists. Since $𝚽$ is a solution to $𝐗\prime =𝐀𝐗$, we have $𝚽\prime =𝐀𝚽$. Suppose ${𝐗}_p=𝚽(t)𝐔(t)$, where $𝐔(t)$ is a column matrix of unknown functions.

Substitute into $𝐗\prime =𝐀𝐗+𝐅$. By the product rule, ${𝐗}_p^{\prime }=𝚽\prime 𝐔+𝚽𝐔\prime$, so

$$\begin{array}{cc}\hfill {\displaystyle 𝚽\prime 𝐔+𝚽𝐔\prime }& {\displaystyle =𝐀𝚽𝐔+𝐅}\hfill \\ \hfill {\displaystyle 𝐀𝚽𝐔+𝚽𝐔\prime }& {\displaystyle =𝐀𝚽𝐔+𝐅}\hfill \end{array}$$

The $𝐀𝚽𝐔$ terms cancel from both sides:

$${\displaystyle 𝚽𝐔\prime =𝐅⟹𝐔\prime ={𝚽}^{-1}𝐅⟹𝐔=\int {𝚽}^{-1}𝐅(t)dt}$$

So the general solution is

$${\displaystyle 𝐗(t)=𝚽(t)𝐂+𝚽(t)\int {𝚽}^{-1}(t)𝐅(t)dt}$$

The first term is ${𝐗}_c$ and the second is ${𝐗}_p$.

8.3c

Example 2

Use variation of parameters to solve the given nonhomogeneous system.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{dx}{dt}}& {\displaystyle =2x-y}\hfill \\ \hfill {\displaystyle \frac{dy}{dt}}& {\displaystyle =3x-2y+4t}\hfill \end{array}}$$

Solution

$$𝐗\prime =\left(\begin{array}{cc}2& -1\\ 3& -2\end{array}\right)𝐗+\left(\begin{array}{c}0\\ 4t\end{array}\right)$$

$$\left|\begin{array}{cc}2-\lambda & -1\\ 3& -2-\lambda \end{array}\right|=0⟹\lambda =1,-1⟹{𝐊}_1=\left(\begin{array}{c}1\\ 1\end{array}\right),{𝐊}_2=\left(\begin{array}{c}1\\ 3\end{array}\right)$$

Each solution becomes a column of $𝚽$:

$$\begin{array}{cc}\hfill {\displaystyle {𝐗}_c}& {\displaystyle =c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^t+c_2\left(\begin{array}{c}1\\ 3\end{array}\right)e^{-t}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}{e}^t\\ e^t\end{array}\right)c_1+\left(\begin{array}{c}{e}^{-t}\\ 3e^{-t}\end{array}\right)c_2=\left(\begin{array}{cc}{e}^t& e^{-t}\\ e^t& 3e^{-t}\end{array}\right)\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)}\hfill \end{array}$$

$$𝚽=\left(\begin{array}{cc}{e}^t& e^{-t}\\ e^t& 3e^{-t}\end{array}\right)$$

Here $det𝚽=3-1=2$:

$${\displaystyle {𝚽}^{-1}=\frac{1}{2}\left(\begin{array}{cc}3e^{-t}& -e^{-t}\\ -e^t& e^t\end{array}\right)=\left(\begin{array}{cc}\frac{3}{2}{e}^{-t}& -\frac{1}{2}{e}^{-t}\\ -\frac{1}{2}{e}^t& \frac{1}{2}{e}^t\end{array}\right)}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \int {𝚽}^{-1}𝐅(t)dt}& {\displaystyle =\frac{1}{2}\int \left(\begin{array}{cc}3e^{-t}& -e^{-t}\\ -e^t& e^t\end{array}\right)\left(\begin{array}{c}0\\ 4t\end{array}\right)dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{2}\int \left(\begin{array}{c}-4t{e}^{-t}\\ 4t{e}^t\end{array}\right)dt=\left(\begin{array}{c}2t{e}^{-t}+2e^{-t}\\ 2t{e}^t-2e^t\end{array}\right)}\hfill \end{array}}$$

Tip: No constant of integration is needed here: we only need one particular solution, and the constants are already in ${𝐗}_c$.

Multiply by $𝚽$, row 1 and then row 2 times the column:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {𝐗}_p=𝚽\int {𝚽}^{-1}𝐅dt}& {\displaystyle =\left(\begin{array}{cc}{e}^t& e^{-t}\\ e^t& 3e^{-t}\end{array}\right)\left(\begin{array}{c}2t{e}^{-t}+2e^{-t}\\ 2t{e}^t-2e^t\end{array}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}2t+2+2t-2\\ 2t+2+6t-6\end{array}\right)=\left(\begin{array}{c}4t\\ 8t-4\end{array}\right)}\hfill \end{array}}$$

Write $\left(\begin{array}{c}4t\\ 8t-4\end{array}\right)=\left(\begin{array}{c}4\\ 8\end{array}\right)t+\left(\begin{array}{c}0\\ -4\end{array}\right)$:

Equivalently, $𝐗=\left(\begin{array}{cc}{e}^t& e^{-t}\\ e^t& 3e^{-t}\end{array}\right)𝐂+\left(\begin{array}{c}4t\\ 8t-4\end{array}\right)$.

8.3d

Example 3

Use variation of parameters to solve the given nonhomogeneous system.

$$𝐗\prime =\left(\begin{array}{cc}2& -1\\ 4& 2\end{array}\right)𝐗+\left(\begin{array}{c}\sin 2t\\ 2\cos 2t\end{array}\right)e^{2t}$$

Solution

The plan: eigenvalues $\lambda$, then eigenvectors $𝐊$, then ${𝐗}_c$ and $𝚽$.

$$\left|\begin{array}{cc}2-\lambda & -1\\ 4& 2-\lambda \end{array}\right|=0⟹(2-\lambda {)}^2=-4⟹\lambda =2\pm 2i$$

For $\lambda =2+2i$:

$$\left(\begin{array}{cc}-2i& -1\\ 4& -2i\end{array}\right)𝐊=\mathbf{0}⟹-2i{k}_1=k_2$$

Let $k_1=1$. Then $k_2=-2i$:

$$𝐊=\left(\begin{array}{c}1\\ -2i\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ -2\end{array}\right)i={𝐁}_1+{𝐁}_{2}i$$

Tip: Choosing $k_1=i$ instead gives $k_2=2$ and $𝐊=\left(\begin{array}{c}0\\ 2\end{array}\right)+\left(\begin{array}{c}1\\ 0\end{array}\right)i$. Any nonzero multiple of an eigenvector works; it leads to an equivalent ${𝐗}_c$.

$${𝐗}_c=c_1{e}^{2t}[\left(\begin{array}{c}1\\ 0\end{array}\right)\cos 2t-\left(\begin{array}{c}0\\ -2\end{array}\right)\sin 2t]+c_2{e}^{2t}[\left(\begin{array}{c}0\\ -2\end{array}\right)\cos 2t+\left(\begin{array}{c}1\\ 0\end{array}\right)\sin 2t]$$

$${𝐗}_c=𝚽𝐂=\left(\begin{array}{cc}{e}^{2t}\cos 2t& e^{2t}\sin 2t\\ 2e^{2t}\sin 2t& -2e^{2t}\cos 2t\end{array}\right)𝐂$$

The determinant is $det𝚽=-2e^{4t}{\cos }^{2}2t-2e^{4t}{\sin }^{2}2t=-2e^{4t}$, so

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {𝚽}^{-1}}& {\displaystyle =-\frac{1}{2e^{4t}}\left(\begin{array}{cc}-2e^{2t}\cos 2t& -e^{2t}\sin 2t\\ -2e^{2t}\sin 2t& e^{2t}\cos 2t\end{array}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-2t}\left(\begin{array}{cc}\cos 2t& \frac{1}{2}\sin 2t\\ \sin 2t& -\frac{1}{2}\cos 2t\end{array}\right)}\hfill \end{array}}$$

In ${𝚽}^{-1}𝐅$, the $e^{-2t}$ from ${𝚽}^{-1}$ cancels the $e^{2t}$ from $𝐅$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \int {𝚽}^{-1}𝐅dt}& {\displaystyle =\int e^{-2t}\left(\begin{array}{cc}\cos 2t& \frac{1}{2}\sin 2t\\ \sin 2t& -\frac{1}{2}\cos 2t\end{array}\right)\left(\begin{array}{c}\sin 2t\\ 2\cos 2t\end{array}\right)e^{2t}dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\int \left(\begin{array}{c}2\sin 2t\cos 2t\\ {\sin }^{2}2t-{\cos }^{2}2t\end{array}\right)dt=\int \left(\begin{array}{c}\sin 4t\\ -\cos 4t\end{array}\right)dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}-\frac{1}{4}\cos 4t\\ -\frac{1}{4}\sin 4t\end{array}\right)}\hfill \end{array}}$$

(The double-angle identities $2\sin \theta \cos \theta =\sin 2\theta$ and ${\cos }^2\theta -{\sin }^2\theta =\cos 2\theta$ turn the entries into $\sin 4t$ and $-\cos 4t$.)

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {𝐗}_p=𝚽\int {𝚽}^{-1}𝐅dt}& {\displaystyle =e^{2t}\left(\begin{array}{cc}\cos 2t& \sin 2t\\ 2\sin 2t& -2\cos 2t\end{array}\right)\left(\begin{array}{c}-\frac{1}{4}\cos 4t\\ -\frac{1}{4}\sin 4t\end{array}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{2t}\left(\begin{array}{c}-\frac{1}{4}(\cos 2t\cos 4t+\sin 2t\sin 4t)\\ \frac{1}{2}(-\sin 2t\cos 4t+\cos 2t\sin 4t)\end{array}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{2t}\left(\begin{array}{c}-\frac{1}{4}\cos 2t\\ \frac{1}{2}\sin 2t\end{array}\right)}\hfill \end{array}}$$

The last step uses the difference formulas: $\cos 4t\cos 2t+\sin 4t\sin 2t=\cos (4t-2t)$ and $\sin 4t\cos 2t-\cos 4t\sin 2t=\sin (4t-2t)$.

8.3e

For an initial-value problem of the form $𝐗\prime =𝐀𝐗+𝐅(t),𝐗(t_0)={𝐗}_0$, we have

$${\displaystyle 𝐗=𝚽𝐂+𝚽{\int }_{t_0}^t{𝚽}^{-1}(s)𝐅(s)ds⟹𝐗(t_0)=𝚽(t_0)𝐂⟹𝐂={𝚽}^{-1}(t_0)𝐗(t_0).}$$

Here $s$ is a dummy variable of integration, so that $t$ can be the upper limit.

Thus,

$${\displaystyle 𝐗(t)=𝚽(t){𝚽}^{-1}(t_0)𝐗(t_0)+𝚽(t){\int }_{t_0}^t{𝚽}^{-1}(s)𝐅(s)ds}$$

Example 4

Solve the given initial-value problem.

$$𝐗\prime =\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)𝐗+\left(\begin{array}{c}1/t\\ 1/t\end{array}\right),𝐗(1)=\left(\begin{array}{c}2\\ -1\end{array}\right)$$

Solution

$$\left|\begin{array}{cc}1-\lambda & -1\\ 1& -1-\lambda \end{array}\right|=0⟹{\lambda }^2=0⟹\lambda =0\text{(repeated)}$$

$$\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)𝐊=\mathbf{0}⟹𝐊=\left(\begin{array}{c}1\\ 1\end{array}\right)$$

The eigenvalue is repeated, so also solve $(𝐀-\lambda 𝐈)𝐏=𝐊$:

$$\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)𝐏=\left(\begin{array}{c}1\\ 1\end{array}\right)⟹p_1-p_2=1$$

Let $p_2=0$. Then $p_1=1$, so $𝐏=\left(\begin{array}{c}1\\ 0\end{array}\right)$. With $\lambda =0$, every $e^{\lambda t}$ is 1:

$${𝐗}_c=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)+c_2[\left(\begin{array}{c}1\\ 1\end{array}\right)t+\left(\begin{array}{c}1\\ 0\end{array}\right)]=\left(\begin{array}{cc}1& t+1\\ 1& t\end{array}\right)𝐂$$

$$𝚽=\left(\begin{array}{cc}1& t+1\\ 1& t\end{array}\right)⟹det𝚽=t-(t+1)=-1$$

$${𝚽}^{-1}=\left(\begin{array}{cc}-t& t+1\\ 1& -1\end{array}\right)⟹{𝚽}^{-1}𝐅=\left(\begin{array}{cc}-t& t+1\\ 1& -1\end{array}\right)\left(\begin{array}{c}1/t\\ 1/t\end{array}\right)=\left(\begin{array}{c}1/t\\ 0\end{array}\right)$$

(Row 1: ${\displaystyle -1+\frac{t+1}{t}=\frac{1}{t}}$. Row 2: ${\displaystyle \frac{1}{t}-\frac{1}{t}=0}$.) Integrate from $t_0=1$ to $t$, using the dummy variable $s$:

$${\displaystyle {\int }_1^t\left(\begin{array}{c}1/s\\ 0\end{array}\right)ds={\left(\begin{array}{c}\ln s\\ 0\end{array}\right)|}_1^t=\left(\begin{array}{c}\ln t\\ 0\end{array}\right)}$$

$${𝐗}_p=𝚽\left(\begin{array}{c}\ln t\\ 0\end{array}\right)=\left(\begin{array}{cc}1& t+1\\ 1& t\end{array}\right)\left(\begin{array}{c}\ln t\\ 0\end{array}\right)=\left(\begin{array}{c}\ln t\\ \ln t\end{array}\right)$$

For the first term, $𝚽(t){𝚽}^{-1}(t_0)𝐗(t_0)$, use ${𝚽}^{-1}(1)=\left(\begin{array}{cc}-1& 2\\ 1& -1\end{array}\right)$:

$$𝐂={𝚽}^{-1}(1)𝐗(1)=\left(\begin{array}{cc}-1& 2\\ 1& -1\end{array}\right)\left(\begin{array}{c}2\\ -1\end{array}\right)=\left(\begin{array}{c}-4\\ 3\end{array}\right)$$

$$\begin{array}{cc}\hfill {\displaystyle 𝐗(t)}& {\displaystyle =𝚽𝐂+{𝐗}_p}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{cc}1& t+1\\ 1& t\end{array}\right)\left(\begin{array}{c}-4\\ 3\end{array}\right)+\left(\begin{array}{c}\ln t\\ \ln t\end{array}\right)}\hfill \end{array}$$