MATH 308 Lecture 26

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Lecture Notes

Systems of Differential Equations

$\{\begin{array}{l}{x_1}^{\prime }(t)=-2x_1+x_2\\ {x_2}^{\prime }(t)=x_1-2x_2\end{array}$

1. solve first equation for $x_2$
2. substitute into the second equation, thereby obtaining a second order equation for $x_1$
3. solve equation for $x_1$
4. determine $x_2$

$\begin{array}{rl}{x}_2& ={x_1}^{\prime }+2x_1\\ \frac{\mathrm{d}}{\mathrm{d}x}({x_1}^{\prime }+2x_1)& =x_1-2({x_1}^{\prime }+2x_1)\\ 0& ={x_1}^{″}+4{x_1}^{\prime }+3x_1\\ x_1& =c_1{\mathrm{e}}^{-t}+c_2{\mathrm{e}}^{-3t}\\ x_2& =(-c_1{\mathrm{e}}^{-t}-3c_2{\mathrm{e}}^{-3t})+2(c_1{\mathrm{e}}^{-t}+c_2{\mathrm{e}}^{-3t})\\ & =c_1{\mathrm{e}}^{-t}-c_2{\mathrm{e}}^{-3t}\end{array}$

Satisfying Initial Conditions

Find a particular solution of the system above that also satisfies the initial conditions $x_1(0)=2$, $x_2(0)=3$.

$\{\begin{array}{l}{x}_1(0)=c_1+c_2=2\\ x_2(0)=c_1-c_2=3\end{array}$

$c_1=\frac{5}{2}$, and $c_2=-\frac{1}{2}$, so the particular solutions are:

$\{\begin{array}{l}{x}_1=\frac{5}{2}{\mathrm{e}}^{-t}-\frac{1}{2}{\mathrm{e}}^{-3t}\\ x_2=\frac{5}{2}{\mathrm{e}}^{-t}+\frac{1}{2}{\mathrm{e}}^{-3t}\end{array}$

Matrix Notation

(See MATH 323 Lecture 26#Matrix Exponential→)

The previous system of equations can be represented as the matrix equation

${\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]}^{\prime }=\left[\begin{array}{cc}-2& 1\\ 1& -2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]$

In general, we now have the equations represented as $X^{\prime }=AX$, and the solution will be $X=\left({\mathrm{e}}^{At}\right)X_0$:

$X^{\prime }=AX⟺X=\left({\mathrm{e}}^{At}\right)X_0$

Exercise 2

Express the system of differential equations in matrix notation:

$\{\begin{array}{l}{x_1}^{\prime }(t)=2t{x}_1(t)+5x_3(t)\\ {x_2}^{\prime }(t)=(\sin t)x_1(t)+t^2{x}_2(t)\\ {x_3}^{\prime }(t)=-x_1(t)+x_2(t)+3x_3(t)\end{array}$

${\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]}^{\prime }=\left[\begin{array}{ccc}2t& 0& 5\\ \sin t& t^2& 0\\ -1& 1& 3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$

Exercise 3

Transform the differential equation into a system of first order equations. Express the system in matrix notation: $y^{″}+3y^{\prime }-5y=\cos 3x$

1. Let $x_1=y$. Then ${x_1}^{\prime }=y^{\prime }$
2. Let $x_2=y^{\prime }$. Then ${x_1}^{\prime }=x_2$ (from above) and ${x_2}^{\prime }=y^{″}=5y-3y^{\prime }+\cos 3x=5x_1-3x_2+\cos 3x$

Therefore, our system is

$\{\begin{array}{l}{x_1}^{\prime }=x_2\\ {x_2}^{\prime }=5x_1-3x_2+\cos 3x\end{array}$

This can be expressed as the matrix equation

${\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]}^{\prime }=\left[\begin{array}{cc}0& 1\\ 5& -3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]+\left[\begin{array}{c}0\\ \cos 3x\end{array}\right]$

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Transform $y^{(4)}+3y^{\prime }=5$ into a system of first order equations.

1. Let $x_1=y$. Then ${x_1}^{\prime }=y^{\prime }$
2. Let $x_2=y^{\prime }$. Then ${x_1}^{\prime }=x_2$ and ${x_2}^{\prime }=y^{″}$
3. Let $x_3=y^{″}$. Then ${x_2}^{\prime }=x_3$ and ${x_3}^{\prime }=y^{(3)}$
4. Let $x_4=y^{(3)}$. Then ${x_3}^{\prime }=x_4$ and ${x_4}^{\prime }=y^{(4)}=-3y^{\prime }+5=-3x_2+5$

Therefore, our system is

$\{\begin{array}{l}{x_1}^{\prime }=x_2\\ {x_2}^{\prime }=x_3\\ {x_3}^{\prime }=x_4\\ {x_4}^{\prime }=-3x_2+5\end{array}$

and can be expressed as

${\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]}^{\prime }=\left[\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& -3& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 0\\ 5\end{array}\right]$