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Differential Equations : Nonhomogeneous Linear Systems

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Example Questions

Example Question #11 : System Of Linear First Order Differential Equations

Solve the following system. $\mathbf{x'} =\begin{bmatrix} -4&2 \\ -1&-1 \end{bmatrix}\mathbf{x} + e^{3t}\begin{bmatrix} 3\\4 \end{bmatrix}$

Possible Answers:

$\mathbf{x} =c_1e^{-2t}\begin{bmatrix} 1\\1 \end{bmatrix} + c_2 e^{-3t}\begin{bmatrix} 2\\1 \end{bmatrix} + e^{3t}\begin{bmatrix} 2\\ \frac{7}{8} \end{bmatrix}$

a $\mathbf{x} =c_1e^{-2t}\begin{bmatrix} 1\\1 \end{bmatrix} + c_2 e^{-3t}\begin{bmatrix} 2\\1 \end{bmatrix} + e^{5t}\begin{bmatrix} \frac{4}{3}\\ \frac{7}{8} \end{bmatrix}$

$\mathbf{x} =c_1e^{2t}\begin{bmatrix} -2\\-1 \end{bmatrix} + c_2 e^{3t}\begin{bmatrix} 1\\-1 \end{bmatrix} + e^{5t}\begin{bmatrix} 1\\ 3 \end{bmatrix}$

$\mathbf{x} =c_1e^{-2t}\begin{bmatrix} 1\\1 \end{bmatrix} + c_2 e^{-3t}\begin{bmatrix} 2\\1 \end{bmatrix} + e^{5t}\begin{bmatrix} 3\\ \frac{7}{8} \end{bmatrix}$

$\mathbf{x} =c_1e^{-2t}\begin{bmatrix} 1\\1 \end{bmatrix} + c_2 e^{-3t}\begin{bmatrix} 2\\1 \end{bmatrix} + e^{3t}\begin{bmatrix} \frac{2}{3}\\ \frac{5}{6} \end{bmatrix}$

Correct answer:

$\mathbf{x} =c_1e^{-2t}\begin{bmatrix} 1\\1 \end{bmatrix} + c_2 e^{-3t}\begin{bmatrix} 2\\1 \end{bmatrix} + e^{3t}\begin{bmatrix} \frac{2}{3}\\ \frac{5}{6} \end{bmatrix}$

Explanation:

First, we will need the complementary solution, and a fundamental matrix for the homogeneous system. Thus, we find the characteristic equation of the matrix given.

$\begin{vmatrix} \lambda+4 &-2 \\ 1& \lambda +1 \end{vmatrix} = (\lambda+4)(\lambda+1) +2 = \lambda^2 + 5\lambda +6 = (\lambda + 3)(\lambda + 2) = 0$

Using $\lambda = -2, -3$ , we then find the eigenvectors by solving for the eigenspace.

$\lambda = -2$

$\begin{bmatrix} 2 & -2 &\vline &0 \\ 1& -1 &\vline & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -1 &\vline &0 \\ 0& 0 &\vline & 0 \end{bmatrix}$

This has solutions x_1 - x_2 = 0 , or $x_1\begin{bmatrix} 1\\ 1 \end{bmatrix}$ . So a suitable eigenvector is simply $\begin{bmatrix} 1\\1 \end{bmatrix}$ .

Repeating for $\lambda = -3$ ,

$\begin{bmatrix} 1 & -2 &\vline &0 \\ 1& -2 &\vline & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -2 &\vline &0 \\ 0& 0 &\vline & 0 \end{bmatrix}$

This has solutions x_1 - 2x_2 = 0 , and thus a suitable eigenvector is $\begin{bmatrix} 2\\1 \end{bmatrix}$ . Thus, our complementary solution is $\mathbf{x} = c_1e^{-2t}\begin{bmatrix} 1\\1 \end{bmatrix} + c_2 e^{-3t}\begin{bmatrix} 2\\1 \end{bmatrix}$ and our fundamental matrix (though in this case, not the matrix exponential) is $X = \begin{bmatrix} e^{-2t} &2e^{-3t} \\ e^{-2t}&e^{-3t} \end{bmatrix}$ . Variation of parameters tells us that the particular solution is given by $\mathbf{x}_p = X\int X^{-1}f(t)dt$ , so first we find $X^{-1}$ using the inverse rule for 2x2 matrices. Thus, $X^{-1} = -\frac{1}{e^{-5t}}\begin{bmatrix} e^{-3t} &-2e^{-3t} \\ -e^{-2t}&e^{-2t} \end{bmatrix} = \begin{bmatrix} -e^{2t} &2e^{2t} \\ e^{3t}&-e^{3t} \end{bmatrix}$ . Plugging in, we have $X^{-1}f(t) = \begin{bmatrix} -e^{2t} &2e^{2t} \\ e^{3t}&-e^{3t} \end{bmatrix}\begin{bmatrix} 3e^{3t}\\4e^{3t} \end{bmatrix} = \begin{bmatrix} 5e^{5t}\\-e^{6t} \end{bmatrix}$ . So $\intX^{-1}f(t)dt = \begin{bmatrix} \int5e^{5t}dt\\ \int-e^{6t}dt \end{bmatrix}=$ $\begin{bmatrix} e^{5t}\\ -\frac{e^{6t}}{6} \end{bmatrix}$ .

Finishing up, we have $\mathbf{x}_p = X\int X^{-1}f(t)dt = \begin{bmatrix} e^{-2t} &2e^{-3t} \\ e^{-2t}&e^{-3t} \end{bmatrix}\begin{bmatrix} e^{5t}\\ -\frac{e^{6t}}{6} \end{bmatrix} = e^{3t}\begin{bmatrix} \frac{2}{3}\\ \frac{5}{6} \end{bmatrix}$ .

Adding the particular solution to the homogeneous, we get a final general solution of $\mathbf{x} =c_1e^{-2t}\begin{bmatrix} 1\\1 \end{bmatrix} + c_2 e^{-3t}\begin{bmatrix} 2\\1 \end{bmatrix} + e^{3t}\begin{bmatrix} \frac{2}{3}\\ \frac{5}{6} \end{bmatrix}$

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Differential Equations : Nonhomogeneous Linear Systems

Study concepts, example questions & explanations for Differential Equations

All Differential Equations Resources

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Example Question #11 : System Of Linear First Order Differential Equations

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