Nonhomogeneous differential equations

Let $A$ be the square matrix of order $n$ . Suppose that ${\bf X}_{1},{\bf X}_{2},\ldots,{\bf X}_{n}$ are linearly independent solutions of ${\bf X}^{\prime} = A{\bf X}$ . Then

$\displaystyle \Phi = \left(\begin{array}{cccc} {\bf X}_{1}&{\bf X}_{2}&\cdots&{\bf X}_{n} \end{array}\right)$

is called the fundamental matrix. Now
$1. \ \det\Phi = W({\bf X}_{1},{\bf X}_{2},\ldots,{\bf X}_{n}) \neq 0 \ (W = \mbox{Wronskian})$
$2. \ \Phi^{\prime} = A\Phi$
$3. \ c_{1}{\bf X}_{1} + c_{2}{\bf X}_{2} + \cdots + c_{n}{\bf X}_{n} = \Phi\lef... ...in{array}{c} c_{1}\\ c_{2}\\ \vdots\\ c_{n} \end{array}\right) = \Phi{\bf C}$
Thus the complementary solution of ${\bf X}^{\prime} = A{\bf X}$ is given by $\Phi{\bf C}$ . Now using the variation of parameter, we let $\Phi(t) {\bf U}(t)$ be the solution of ${\bf X}^{\prime} = A{\bf X} + {\bf F}$ . Then since the derivative of the vector valued function is given by the derivatives of components of the vector valued function, we have

$\displaystyle (\Phi {\bf U})^{\prime} = \Phi^{\prime} {\bf U} + \Phi {\bf U}^{\prime}$

Now substitute ${\bf X} = \Phi {\bf U}$ into ${\bf X}^{\prime} = A{\bf X} + {\bf F}$ . Then

$\displaystyle \Phi^{\prime} {\bf U} + \Phi {\bf U}^{\prime} = A\Phi {\bf U} + {\bf F}$

Since $\Phi^{\prime} = A\Phi$ , we have

$\displaystyle \Phi {\bf U}^{\prime} = {\bf F}$

Then by the Cramer's rule

$\displaystyle {\bf U}^{\prime} = \frac{[\Phi:{\bf F}]}{\vert\Phi\vert}$

Integrate to get the general solution

$\displaystyle {\bf X} = \Phi {\bf C} + \Phi {\bf U}$

Example 3..5 Solve the differential equation ${\bf X}^{\prime} = \left(\begin{array}{rr} 1&2\\ 3&2 \end{array}\right){\bf X} + \left(\begin{array}{c} e^{3t}\\ 2e^{3t} \end{array}\right)$

SOLUTION $\det(A - \lambda I) = \lambda^2 -3\lambda -4 = (\lambda +1)(\lambda -4) = 0$ . Thus the eigenvalues are $\lambda = -1,4$ . The eigenvector corresponds to $\lambda = -1$ is obtained by $(A - (-1(I) = \left(\begin{array}{rr} 2&2\\ 3&3 \end{array}\right)$ . Then the eigenvector is $\left(\begin{array}{c} -1\\ 1 \end{array}\right)$ . Thus ${\bf X}_{1} = \left(\begin{array}{c} -1\\ 1 \end{array}\right)e^{-t}$ is a solution. The eigenvector corresponds to $\lambda = 4$ is obtained by $(A - 4I) = \left(\begin{array}{rr} -3&2\\ 3&-2 \end{array}\right)$ . Thus the eigenvector is $\left(\begin{array}{c} 2\\ 3 \end{array}\right)$ . Therefore, ${\bf X}_{2} = \left(\begin{array}{c} 2\\ 3 \end{array}\right)e^{4t}$ is a solution. Then the fundamental matrix is

$\displaystyle \Phi(t) = \left(\begin{array}{rr} -e^{-t}&2e^{4t}\\ e^{-t}&3e^{4t} \end{array}\right)$

. Now to find the general solution, we need to solve

$\displaystyle \Phi {\bf U}^{\prime} = {\bf F}$

$\displaystyle \left(\begin{array}{rr} -e^{-t}&2e^{4t}\\ e^{-t}&3e^{4t} \end{ar... ...nd{array}\right) = \left(\begin{array}{c} e^{3t}\\ 2e^{3t} \end{array}\right)$

By Cramer's rule,

$\displaystyle u_{1}^{\prime} = \frac{\left\vert\begin{array}{rr} e^{3t}&2e^{4t}... ...3e^{4t} \end{array}\right\vert} = \frac{-e^{7t}}{-5e^{3t}} = \frac{1}{5}e^{4t}$

$\displaystyle u_{2}^{\prime} = \frac{\left\vert\begin{array}{rr} -e^{-t}&e^{3t}... ...d{array}\right\vert}{-5e^{3t}} = \frac{-3e^{2t}}{-5e^{3t}} = \frac{3}{5}e^{-t}$

Integrating,

$\displaystyle u_{1} = \frac{1}{20}e^{4t}, \ u_{2} = -\frac{3}{5}e^{-t}$

Thus the general solution is

$\displaystyle {\bf X}$	$\displaystyle =$	$\displaystyle \Phi {\bf C} + \Phi {\bf U} = \left(\begin{array}{rr} -e^{-t}&2e^... ...ft(\begin{array}{c} \frac{1}{20}e^{4t}\\ -\frac{3}{5}e^{-t} \end{array}\right)$
	$\displaystyle =$	$\displaystyle \left(\begin{array}{l} -c_{1}e^{-t}+ 2c_{2}e^{4t} - \frac{5}{4}e^{3t}\\ c_{1}e^{-t}+ 3c_{2}e^{4t} - \frac{7}{4}e^{3t} \end{array}\right)$

$\ \blacksquare$

Let $D = d/dt$ . Then $x_{1}^{\prime} = Dx_{1}$ and the given differential equation

$\displaystyle {\bf X}^{\prime} = \left(\begin{array}{rr} 1&2\\ 3&2 \end{array}\right){\bf X} + \left(\begin{array}{c} e^{3t}\\ 2e^{3t} \end{array}\right)$

can be written as

$\displaystyle \left\{\begin{array}{rl} (D - 1)x_{1} - 2x_{2} &= e^{3t}\\ -3x_{1} + (D - 2)x_{2} &= 2e^{3t} \end{array}\right.$

Then solve for $x_1$

by using Cramer's rule.

Example 3..6 Solve the following differential equation

$\displaystyle \left\{\begin{array}{rl} (D - 1)x_{1} - 2x_{2} &= e^{3t}\\ -3x_{1} + (D - 2)x_{2} &= 2e^{3t} \end{array}\right.$

(3.1)

SOLUTION Using Cramer's rule, we have

$\displaystyle \left\vert\begin{array}{cc} D-1 & -2\\ -3 & D-2 \end{array}\righ... ...\vert\begin{array}{cc} e^{3t} & -2\\ 2e^{3t} & D - 2 \end{array}\right\vert .$

Then

$\displaystyle (D^{2} - 3D - 4)x_{1} = (D-2)e^{3t} + 2(2e^{3t}) = 5e^{3t}.$

The characteristic equation of the equation is $m^2 - 3m - 4 = 0$

. Thus

and the complementary solution $x_{1c}$ is

$\displaystyle x_{1c} = c_{1}e^{-t} + c_{2} e^{4t}.$

To find $x_{1p}$ , we use the method of undetermined coefficients

$\displaystyle H(D)5e^{3t} = (D-3)5e^{3t} = 0$

implies that

$\displaystyle (D-3)(D^2 - 3D - 4)x_{1} = (D-3)5e^{3t} = 0.$

Thus, $x_{1p} = Ae^{3t}$ .

$\displaystyle 9Ae^{3t} - 9Ae^{3t} - 4Ae^{3t} = 5e^{3t}.$

Solve this to get $A = -\frac{5}{4}$ . Then

$\displaystyle x_{1} = x_{1c} + x_{1p} = c_{1}e^{-t} + c_{2}e^{4t} - \frac{5}{4}e^{3t}.$

Now put this back to (3.1).

$\displaystyle -c_{1}e^{-t} + 4c_{2}e^{4t} - \frac{15}{4}e^{3t} - c_{1}e^{-t} - c_{2}e^{4t} + \frac{5}{4}e^{3t} - 2x_{2} = e^{3t}.$

Thus,

$\displaystyle x_{2} = -c_{1}e^{-t} + \frac{3}{2}c_{2}e^{4t} - \frac{7}{4}e^{2t}\ensuremath{\ \blacksquare}$

Example 3..7 Solve the following differential equation

$\displaystyle \left\{\begin{array}{rc} x_{1}^{\prime\prime} - 2x_{1} - 3x_{2} =& 0\\ x_{1} + x_{2}^{\prime\prime} + 2x_{2} =& 0 \end{array}\right .$

SOLUTION Let $u = x_{1}^{\prime}$ and $v = x_{2}^{\prime}$ . Then $u^{\prime} = x_{1}^{\prime\prime}$ and $v^{\prime} = x_{2}^{\prime\prime}$ . Thus we can express

$\displaystyle \left(\begin{array}{c} x_{1}\\ x_{2}\\ u\\ v \end{array}\right... ...rray}\right)\left(\begin{array}{c} x_{1}\\ x_{2}\\ u\\ v \end{array}\right)$

$A = \left(\begin{array}{rrrr} 0&0&1&0\\ 0&0&0&1\\ 2&3&0&0\\ -1&-2&0&0 \end{array}\right)$ Then the eigeneqution is $\lambda^{4} - 1 = 0$ and the eigenvalues are $\lambda = \pm 1, \pm i$ . To find the eigenvector, we use Mathematica. Then the eigenvectors correspond to $\lambda = 1,-1,i$ are

$\displaystyle (3,-1,-3,1)^{t},(-3,1,-3,1)^{t},(-i,i,-1,1)^{t}$

Thus ${\bf C}e^{\lambda t}$ is

$\displaystyle \left(\begin{array}{c} 3e^{t}\\ -e^{t}\\ -3e^{t}\\ e^{t} \end{... ...\cos{t}+i\sin{t})\\ -(\cos{t}+i\sin{t})\\ \cos{t}+i\sin{t} \end{array}\right)$

Thus the fundamental matric $\Phi(t)$ is

$\displaystyle \Phi(t) = \left(\begin{array}{rrrr} 3e^{t}&-3e^{-t}&\sin{t}&-\cos... ...t}&-3e^{-t}&-\cos{t}&\sin{t}\\ e^{t}&e^{t}&\cos{t}&\sin{t} \end{array}\right)$

Therefore the general solution is

$\displaystyle {\bf X} = \left(\begin{array}{rrrr} 3e^{t}&-3e^{-t}&\sin{t}&-\cos... ...ght)\left(\begin{array}{c} c_{1}\\ c_{2}\\ c_{3}\\ c_{4} \end{array}\right)$

$\displaystyle x_{1}$	$\displaystyle =$	$\displaystyle c_{1}3e^{t} -c_{2}3e^{t} = c_{3}\sin{t} - c_{4}\cos{t}$
$\displaystyle x_{2}$	$\displaystyle =$	$\displaystyle -c_{1}e^{t} + c_{2}e^{-t} - c_{3}\sin{t} + c_{4}\cos{t} \ensuremath{\ \blacksquare}$

Subsections

Exercise