MATH 308 Lecture 26

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Lecture Notes

Systems of Differential Equations

${\begin{cases}x_{1}'(t)=-2x_{1}+x_{2}\\x_{2}'(t)=x_{1}-2x_{2}\end{cases}}$

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 $x_{2}$

${\begin{aligned}x_{2}&=x_{1}'+2x_{1}\\{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x_{1}'+2x_{1}\right)&=x_{1}-2\left(x_{1}'+2x_{1}\right)\\0&=x_{1}''+4x_{1}'+3x_{1}\\x_{1}&=c_{1}\,\mathrm {e} ^{-t}+c_{2}\,\mathrm {e} ^{-3t}\\x_{2}&=\left(-c_{1}\,\mathrm {e} ^{-t}-3c_{2}\,\mathrm {e} ^{-3t}\right)+2\left(c_{1}\,\mathrm {e} ^{-t}+c_{2}\,\mathrm {e} ^{-3t}\right)\\&=c_{1}\,\mathrm {e} ^{-t}-c_{2}\,\mathrm {e} ^{-3t}\end{aligned}}$

Satisfying Initial Conditions

Find a particular solution of the system above that also satisfies the initial conditions $x_{1}(0)=2$, $x_{2}(0)=3$.

${\begin{cases}x_{1}(0)=c_{1}+c_{2}=2\\x_{2}(0)=c_{1}-c_{2}=3\end{cases}}$

$c_{1}={\tfrac {5}{2}}$, and $c_{2}=-{\tfrac {1}{2}}$, so the particular solutions are:

${\begin{cases}x_{1}={\frac {5}{2}}\,\mathrm {e} ^{-t}-{\frac {1}{2}}\,\mathrm {e} ^{-3t}\\x_{2}={\frac {5}{2}}\,\mathrm {e} ^{-t}+{\frac {1}{2}}\,\mathrm {e} ^{-3t}\end{cases}}$

Matrix Notation

(See MATH 323 Lecture 26#Matrix Exponential→)

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

${\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}'={\begin{bmatrix}-2&1\\1&-2\end{bmatrix}}\,{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}$

In general, we now have the equations represented as $X'=A\,X$, and the solution will be $X=\left(\mathrm {e} ^{A\,t}\right)\,X_{0}$:

$X'=A\,X\quad \iff \quad X=\left(\mathrm {e} ^{A\,t}\right)\,X_{0}$

Exercise 2

Express the system of differential equations in matrix notation:

${\begin{cases}x_{1}'(t)=2t\,x_{1}(t)+5x_{3}(t)\\x_{2}'(t)=\left(\sin {t}\right)\,x_{1}(t)+t^{2}\,x_{2}(t)\\x_{3}'(t)=-x_{1}(t)+x_{2}(t)+3x_{3}(t)\end{cases}}$

${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}'={\begin{bmatrix}2t&0&5\\\sin {t}&t^{2}&0\\-1&1&3\end{bmatrix}}\,{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}$

Exercise 3

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

1. Let $x_{1}=y$. Then $x_{1}'=y'$
2. Let $x_{2}=y'$. Then $x_{1}'=x_{2}$ (from above) and $x_{2}'=y''=5y-3y'+\cos {3x}=5x_{1}-3x_{2}+\cos {3x}$

Therefore, our system is

${\begin{cases}x_{1}'=x_{2}\\x_{2}'=5x_{1}-3x_{2}+\cos {3x}\end{cases}}$

This can be expressed as the matrix equation

${\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}'={\begin{bmatrix}0&1\\5&-3\end{bmatrix}}\,{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}+{\begin{bmatrix}0\\\cos {3x}\end{bmatrix}}$

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

1. Let $x_{1}=y$. Then $x_{1}'=y'$
2. Let $x_{2}=y'$. Then $x_{1}'=x_{2}$ and $x_{2}'=y''$
3. Let $x_{3}=y''$. Then $x_{2}'=x_{3}$ and $x_{3}'=y^{(3)}$
4. Let $x_{4}=y^{(3)}$. Then $x_{3}'=x_{4}$ and $x_{4}'=y^{(4)}=-3y'+5=-3x_{2}+5$

Therefore, our system is

${\begin{cases}x_{1}'=x_{2}\\x_{2}'=x_{3}\\x_{3}'=x_{4}\\x_{4}'=-3x_{2}+5\end{cases}}$

and can be expressed as

${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}}'={\begin{bmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&-3&0&0\end{bmatrix}}\,{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}}+{\begin{bmatrix}0\\0\\0\\5\end{bmatrix}}$