8.05: On Solving Higher-Order and Coupled Ordinary Differential Equations
- Page ID
- 126431
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Lesson 1: Solving Higher-Order Ordinary Differential Equations
Learning Objectives
After successful completion of this lesson, you should be able to:
1) write higher-order ordinary differential equations as simultaneous first-order ordinary differential equations,
2) solve higher-order ordinary differential equations numerically
Description
In the earlier lessons, we have learned Euler’s and Runge-Kutta methods to solve first-order ordinary differential equations of the form
\[\frac{dy}{dx} = f\left( x,y \right),y\left( x_{0} \right) = y_{0}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.1}) \nonumber\]
What do we do to solve simultaneous (coupled) differential equations or differential equations higher than first order? For example, an \(n^{th}\)order differential equation of the form
\[a_{n}\frac{d^{n}y}{dx^{n}} + a_{n - 1}\frac{d^{n - 1}y}{dx^{n - 1}} + \ldots + a_{1}\frac{dy}{dx} + a_{o}y = f\left( x \right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.2}) \nonumber\]
with \(n\) initial conditions can be solved by assuming
\[y = z_{1}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.3.1}) \nonumber\]
\[\frac{dy}{dx} = \frac{dz_{1}}{dx} = z_{2}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.3.2}) \nonumber\]
\[\frac{d^{2}y}{dx^{2}} = \frac{dz_{2}}{dx} = z_{3}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.3.3}) \nonumber\]
\[\vdots \nonumber\]
\[\frac{d^{n - 1}y}{dx^{n - 1}} = \frac{dz_{n - 1}}{dx} = z_{n}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.3.}n) \nonumber\]
\[\begin{split} \frac{d^{n}y}{dx^{n}} &= \frac{dz_{n}}{dx}\\ &= \frac{1}{a_{n}}\left( - a_{n - 1}\frac{d^{n - 1}y}{dx^{n - 1}}\ldots - a_{1}\frac{dy}{dx} - a_{0}y + f\left( x \right) \right)\\ &= \frac{1}{a_{n}}\left( - a_{n - 1}z_{n}\ldots - a_{1}z_{2} - a_{0}z_{1} + f\left( x \right) \right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.3.}n+1) \end{split}\]
The above Equations from \((\PageIndex{1.3.2})\) to \((\PageIndex{1.3.}n+1)\) represent \(n\) first-order differential equations as follows
\[\frac{dz_{1}}{dx} = z_{2} = f_{1}\left( z_{1},z_{2},\ldots,x \right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.4.1}) \nonumber\]
\[\frac{dz_{2}}{dx} = z_{3} = f_{2}\left( z_{1},z_{2},\ldots,x \right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.4.2}) \nonumber\]
\[\vdots \nonumber\]
\[\frac{dz_{n}}{dx} = \frac{1}{a_{n}}\left( - a_{n - 1}z_{n}\ldots - a_{1}z_{2} - a_{0}z_{1} + f\left( x \right) \right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.4.n}) \nonumber\]
Each of the \(n\) first-order ordinary differential equations should be accompanied by one initial condition. The initial condition should be on the corresponding dependent variable on the left-hand side of the ordinary differential equation. For example, Equation \((\PageIndex{1.4.1})\) would need an initial condition on \(z_{1}\), Equation \((\PageIndex{1.4.}n)\) would need an initial condition on \(z_{n}\), etc. These first-order ordinary differential equations (Equations \((\PageIndex{1.4.1})\) through \((\PageIndex{1.4.}n)\)) are simultaneous. Still, they can be solved by the methods used for solving first-order ordinary differential equations that we have already learned in the previous lessons.
Rewrite the following differential equation as a set of simultaneous first-order differential equations.
\[3\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} + 5y = e^{- x},y\left( 0 \right) = 5,\ y^{\prime}\left( 0 \right) = 7 \nonumber\]
Solution
The ordinary differential equation
\[3\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} + 5y = e^{- x},y\left( 0 \right) = 5,\ y^{\prime}\left( 0 \right) = 7 \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E1.1}) \nonumber\]
would be rewritten as follows. Assume
\[\frac{dy}{dx} = z, \;\;\;\;\;\;\;\;\;\;\;\;(\PageIndex{1.E1.2}) \nonumber\]
Then
\[\frac{d^{2}y}{dx^{2}} = \frac{dz}{dx}\;\;\;\;\;\;\;\;\;\;\;\;(\PageIndex{1.E1.3}) \nonumber\]
Substituting Equations \((\PageIndex{1.E1.2})\) and \((\PageIndex{1.E1.3})\) in the given second-order ordinary differential equation gives
\[3\frac{dz}{dx} + 2z + 5y = e^{- x} \nonumber\]
and rewritten as
\[\frac{dz}{dx} = \frac{1}{3}\left( e^{- x} - 2z - 5y \right) \;\;\;\;\;\;\;\;\;\;\;\;(\PageIndex{1.E1.4}) \nonumber\]
The set of two simultaneous first-order ordinary differential equations complete with the initial conditions then is
\[\frac{dy}{dx} = z,y\left( 0 \right) = 5 \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E1.5}\text{a}) \nonumber\]
\[\frac{dz}{dx} = \frac{1}{3}\left( e^{- x} - 2z - 5y \right), \ z\left( 0 \right) = 7. \;\;\;\;\;\;\;\;\;\;\;\;(\PageIndex{1.E1.5}\text{b}) \nonumber\]
Now one can apply any of the numerical methods used for solving first-order ordinary differential equations.
We write such equations in a subscripted format in a later lesson to make them suitable for matrix and vector representation. In the above example, we would introduce subscripted dependent variables, say \(w_1\) and \(w_2\), as
\[w_1=y \;\;\;\;\;\;\;\;\;\;\;\;(\PageIndex{1.E1.6}) \nonumber\]
and
\[w_2=\displaystyle \frac{dy}{dx}\;\;\;\;\;\;\;\;\;\;\;\;(\PageIndex{1.E1.7}) \nonumber\]
Using Equations \((\PageIndex{1.E1.6})\) and \((\PageIndex{1.E1.7})\), Equations \((\PageIndex{1.E1.5}\text{a})\) and \((\PageIndex{1.E1.5}\text{b})\) can be rewritten in subscripted form as
\[\frac{dw_{1}}{dx} = w_{2}, \ w_{1}\left( 0 \right) = 5 \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E1.8}\text{a}) \nonumber\] \[\frac{dw_{2}}{dx} = \frac{1}{3}\left( e^{- x} - 2w_{2} - 5w_{1} \right), \ w_{2}\left( 0 \right) = 7. \;\;\;\;\;\;\;\;\;\;\;\;(\PageIndex{1.E1.8}\text{b}) \nonumber\]
Rewriting Equations \((\PageIndex{1.E1.8}\text{a})\) and \((\PageIndex{1.E1.8}\text{b})\) now as
\[\frac{dw_{1}}{dx} = 0w_{1}+1w_{2}, \ w_{1}\left( 0 \right) = 5 \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E1.9}\text{a}) \nonumber\] \[\frac{dw_{2}}{dx} = -\frac{5}{3}w_{1} - \frac{2}{3}w_{2} + \frac{1}{3}e^{- x}, \ w_{2}\left( 0 \right) = 7. \;\;\;\;\;\;\;\;\;\;\;\;(\PageIndex{1.E1.9}\text{b}) \nonumber\]
makes it ready to be written in a matrix form and is called the state-space model.
In the matrix form, the Equations \((\PageIndex{1.E1.9}\text{a})\) and \((\PageIndex{1.E1.9}\text{b})\) are given as
\[\begin{bmatrix} \displaystyle \frac{dw_{1}}{dx} \\ \displaystyle\frac{dw_{2}}{dx} \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\displaystyle \frac{5}{3} & -\displaystyle \frac{2}{3} \\ \end{bmatrix}\begin{bmatrix} w_{1} \\ w_{2} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ \displaystyle\frac{1}{3}e^{-x} \\ \end{bmatrix} \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E1.10}) \nonumber\]
where the initial conditions are given by
\[\begin{bmatrix} w_{1}(0) \\ w_{2}(0) \\ \end{bmatrix} = \begin{bmatrix} 5 \\ 7 \\ \end{bmatrix}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E1.11}) \nonumber\]
The motivation and description of the state-space model are described in detail in the next lesson for higher-order and coupled ordinary differential equations.
Given
\[\frac{d^{2}y}{dt^{2}} + 2\frac{dy}{dt} + y = e^{- t}, \ y\left( 0 \right) = 1,\frac{dy}{dt}\left( 0 \right) = 2, \nonumber\]
estimate the following by Euler’s method
a) \(y\left( 0.75 \right)\)
b) the absolute relative true error for part(a), if \(\left. \ y\left( 0.75 \right) \right|_{exact} = 1.668\)
c) \(\displaystyle \frac{dy}{dt}\left( 0.75 \right)\)
Use a step size of \(h = 0.25\).
Solution
First, the second-order differential equation is written as two simultaneous first-order differential equations as follows. Assume
\[\frac{dy}{dt} = z \nonumber\]
then
\[\frac{dz}{dt} + 2z + y = e^{- t} \nonumber\]
\[\frac{dz}{dt} = e^{- t} - 2z - y \nonumber\]
So the two simultaneous first-order differential equations are
\[\frac{dy}{dt} = z = f_{1}\left( t,y,z \right), \ y(0) = 1\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E2.1}) \nonumber\]
\[\frac{dz}{dt} = e^{- t} - 2z - y = f_{2}\left( t,y,z \right),\ z(0) = 2\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E2.2}) \nonumber\]
Using Euler’s method on Equations \((\PageIndex{1.E2.1})\) and \((\PageIndex{1.E2.2})\), we get
\[y_{i + 1} = y_{i} + f_{1}\left( t_{i},y_{i},z_{i} \right)h\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E2.3}) \nonumber\]
\[z_{i + 1} = z_{i} + f_{2}\left( t_{i},y_{i},z_{i} \right)h\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E2.4}) \nonumber\]
a) To find the value of \(y\left( 0.75 \right)\) and since we are using a step size of \(0.25\) and starting at \(t = 0\), we need to take three steps to find the value of \(y\left( 0.75 \right)\).
For \(i = 0, \ t_{0} = 0, \ y_{0} = 1, \ z_{0} = 2\):
From Equation \((\PageIndex{1.E2.3})\)
\[\begin{split} y_{1} &= y_{0} + f_{1}\left( t_{0},y_{0},z_{0} \right)h\\ &= 1 + f_{1}\left( 0,1,2 \right)\left( 0.25 \right)\\ &=1+2\left(0.25\right)\\ &= 1.5 \end{split}\]
\(y_{1}\) is the approximate value of \(y\) at
\[t = t_{1} = t_{0} + h = 0 + 0.25 = 0.25 \nonumber\]
\[y_{1} = y\left( 0.25 \right) \approx 1.5 \nonumber\]
From Equation \((\PageIndex{1.E2.4})\)
\[\begin{split} z_{1} &= z_{0} + f_{2}\left( t_{0},y_{0},z_{0} \right)h\\ &= 2 + f_{2}\left( 0,1,2 \right)\left( 0.25 \right)\\ &= 2 + \left( e^{- 0} - 2\left( 2 \right) - 1 \right)\left( 0.25 \right)\\ &= 1 \end{split}\]
\(z_{1}\) is the approximate value of \(z\) (same as \(\frac{dy}{dt}\)) at \(t = 0.25\)
\[z_{1} = z\left( 0.25 \right) \approx 1 \nonumber\]
For \(i = 1, \ t_{1} = 0.25, \ y_{1} = 1.5, \ z_{1} = 1\):
From Equation \((\PageIndex{1.E2.3})\)
\[\begin{split} y_{2}\ &= y_{1} + f_{1}\left( t_{1},y_{1},z_{1} \right)h\\ &= 1.5 + f_{1}\left( 0.25,1.5,1 \right)\left( 0.25 \right)\\ &= 1.5 + \left( 1 \right)\left( 0.25 \right)\\ &= 1.75 \end{split}\]
\(y_{2}\) is the approximate value of \(y\) at
\[t = t_{2} = t_{1} + h = 0.25 + 0.25 = 0.50 \nonumber\]
\[y_{2} = y\left( 0.5 \right) \approx 1.75 \nonumber\]
From Equation \((\PageIndex{1.E2.4})\)
\[\begin{split} z_{2} &= z_{1} + f_{2}\left( t_{1},y_{1},z_{1} \right)h\\ &= 1 + f_{2}\left( 0.25,1.5,1 \right)\left( 0.25 \right)\\ &= 1 + \left( e^{- 0.25} - 2\left( 1 \right) - 1.5 \right)\left( 0.25 \right)\\ &= 1 + \left( - 2.7211 \right)\left( 0.25 \right)\\ &= 0.31970 \end{split}\]
\(z_{2}\) is the approximate value of \(z\) at
\[t = t_{2} = 0.5 \nonumber\]
\[z_{2} = z\left( 0.5 \right) \approx 0.31970 \nonumber\]
For \(i = 2, \ t_{2} = 0.5, \ y_{2} = 1.75, \ z_{2} = 0.31970\):
From Equation \((\PageIndex{1.E2.3})\)
\[\begin{split} y_{3} &= y_{2} + f_{1}\left( t_{2},y_{2},z_{2} \right)h\\ &= 1.75 + f_{1}\left( 0.50,1.75,0.31970 \right)\left( 0.25 \right)\\ &= 1.75 + \left( 0.31970 \right)\left( 0.25 \right)\\ &= 1.8299 \end{split}\]
\(y_{3}\) is the approximate value of \(y\) at
\[t = t_{3} = t_{2} + h = 0.5 + 0.25 = 0.75 \nonumber\]
\[y_{3} = y\left( 0.75 \right) \approx 1.8299 \nonumber\]
From Equation \((\PageIndex{1.E2.4})\)
\[\begin{split} z_{3} &= z_{2} + f_{2}\left( t_{2},y_{2},z_{2} \right)h\\ &= 0.31972 + f_{2}\left( 0.50,1.75,0.31970 \right)\left( 0.25 \right)\\ &= 0.31972 + \left( e^{- 0.50} - 2\left( 0.31970 \right) - 1.75 \right)\left( 0.25 \right)\\ &= 0.31972 + \left( - 1.7829 \right)\left( 0.25 \right)\\ &= - 0.1260 \end{split}\]
\(z_{3}\) is the approximate value of \(z\) at
\[t = t_{3} = 0.75 \nonumber\]
\[z_{3} = z\left( 0.75 \right) \approx - 0.12601 \nonumber\]
\[y\left( 0.75 \right) \approx y_{3} = 1.8299 \nonumber\]
b) The exact value of \(y\left( 0.75 \right)\) is
\[\left. \ y\left( 0.75 \right) \right|_{exact} = 1.668 \nonumber\]
The absolute relative true error in the result from part (a) is
\[\begin{split} \left| \epsilon_{t} \right| &= \left| \frac{1.668 - 1.8299}{1.668} \right| \times 100\\ &= 9.7062\% \end{split}\]
c)
\[\begin{split} \frac{dy}{dx}\left(0.75\right) &=z_3\\ &\approx - 0.12601 \end{split}\]
Given
\[\frac{d^{2}y}{dt^{2}} + 2\frac{dy}{dt} + y = e^{- t},\ y(0) = 1,\ \frac{dy}{dt}(0) = 2, \nonumber\]
estimate the following by Heun’s method.
a) \(y\left( 0.75 \right)\)
b) \(\displaystyle \frac{dy}{dx}\left( 0.75 \right)\).
Use a step size of \(h = 0.25\).
Solution
First, the second-order differential equation is rewritten as two simultaneous first-order differential equations as follows. Assume
\[\frac{dy}{dt} = z \nonumber\]
then
\[\frac{dz}{dt} + 2z + y = e^{- t} \nonumber\]
\[\frac{dz}{dt} = e^{- t} - 2z - y \nonumber\]
So the two simultaneous first-order differential equations with the corresponding initial conditions are
\[\frac{dy}{dt} = z = f_{1}\left( t,y,z \right), \ y(0) = 1\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E3.1}) \nonumber\]
\[\frac{dz}{dt} = e^{- t} - 2z - y = f_{2}\left( t,y,z \right),\ z(0) = 2\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E3.2}) \nonumber\]
Using Heun’s method on Equations \((\PageIndex{1.E3.1})\) and \((\PageIndex{1.E3.2})\), we get
\[y_{i + 1} = y_{i}\ + \ \frac{1}{2}\left( k_{1}^{y} + k_{2}^{y} \right)h\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E3.3}) \nonumber\]
\[k_{1}^{y} = f_{1}\ \left( t_{i},\ y_{i},\ z_{i} \right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E3.4}\text{a}) \nonumber\]
\[k_{2}^{y} = f_{1}\ \left( t_{i}\ + \ h,\ y_{i}\ + \ hk_{1}^{y},\ z_{i}\ + \ hk_{1}^{z} \right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E3.4}\text{b}) \nonumber\]
\[z_{i + 1} = z_{i}\ + \frac{1}{2}\ \left( k_{1}^{z} + k_{2}^{z} \right)h\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E3.5}) \nonumber\]
\[k_{1}^{z} = \ f_{2}\ \left( t_{i},\ y_{i},\ z_{i} \right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E3.6}\text{a}) \nonumber\]
\[k_{2}^{z} = f_{2}\ \left( t_{i}\ + \ h,\ y_{i}\ + \ hk_{1}^{y},\ z_{i}\ + \ hk_{1}^{z} \right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E3.6}\text{b}) \nonumber\]
For \(i = 0, \ t_{o} = 0, \ y_{o} = 1, \ z_{o} = 2\)
From Equation \((\PageIndex{1.E3.4}\text{a})\)
\[\begin{split} k_{1}^{y} &= f_{1}\left( t_{0},y_{0},z_{0} \right)\\ &= f_{1}\left( 0,1,2 \right)\\ &= 2 \end{split}\]
From Equation \((\PageIndex{1.E3.6}\text{a})\)
\[\begin{split} k_{1}^{z} &= f_{2}\left( t_{0},y_{0},z_{0} \right)\\ &= f_{2}\left( 0,1,2 \right)\\ &= e^{- 0} - 2\left( 2 \right) - 1\\ &= -4 \end{split}\]
From Equation \((\PageIndex{1.E3.4}\text{b})\)
\[\begin{split} k_{2}^{y} &= f_{1}\left( t_{0} + h,y_{0} + hk_{1}^{y},z_{0} + hk_{1}^{z} \right)\\ &= f_{1}\left( 0 + 0.25,1 + \left( 0.25 \right)\left( 2 \right),2 + \left( 0.25 \right)\left( - 4 \right) \right)\\ &= f_{1}\left( 0.25,1.5,1 \right)\\ &= 1 \end{split}\]
From Equation \((\PageIndex{1.E3.6}\text{b})\)
\[\begin{split} k_{2}^{z} &= f_{2}\left( t_{0} + h,y_{0} + hk_{1}^{y},z_{0} + hk_{1}^{z} \right)\\ &= f_{2}\left( 0 + 0.25,1 + \left( 0.25 \right)\left( 2 \right),2 + \left( 0.25 \right)\left( - 4 \right) \right)\\ &= f_{2}\left( 0.25,1.5,1 \right)\\ &= e^{- 0.25} - 2\left( 1 \right) - 1.5\\ &= - 2.7212 \end{split}\]
From Equation \((\PageIndex{1.E3.3})\)
\[\begin{split} y_{1} &= y_{0} + \frac{1}{2}\left( k_{1}^{y} + k_{2}^{y} \right)h\\ &= 1 + \frac{1}{2}\left( 2 + 1 \right)\left( 0.25 \right)\\ &= 1.375 \end{split}\]
\(y_{1}\) is the approximate value of \(y\) at
\[t = t_{1} = t_{0} + h = 0 + 0.25 = 0.25 \nonumber\]
\[y_{1} = y\left( 0.25 \right) \cong 1.375 \nonumber\]
From Equation \((\PageIndex{1.E3.5})\)
\[\begin{split} z_{1} &= z_{0} + \frac{1}{2}\left( k_{1}^{z} + k_{2}^{z} \right)h\\ &= 2 + \frac{1}{2}( - 4 + ( - 2.7212))(0.25)\\ &= 1.1598 \end{split}\]
\(z_{1}\) is the approximate value of \(z\) at
\[t = t_{1} = 0.25 \nonumber\]
\[z_{1} = z\left( 0.25 \right) \approx 1.1598 \nonumber\]
For \(i = 1, \ t_{1} = 0.25, \ y_{1} = 1.375, \ z_{1} = 1.1598\)
From Equation \((\PageIndex{1.E3.4}\text{a})\)
\[\begin{split} k_{1}^{y} &= f_{1}\left( t_{1},y_{1},z_{1} \right)\\ &= f_{1}\left( 0.25,1.375,1.1598 \right)\\ &= 1.1598 \end{split}\]
From Equation \((\PageIndex{1.E3.6}\text{a})\)
\[\begin{split} k_{1}^{z} &= f_{2}\left( t_{1},y_{1},z_{1} \right)\\ &= f_{2}\left( 0.25,1.375,1.1598 \right)\\ &= e^{- 0.25} - 2\left( 1.1598 \right) - 1.375\\ &= - 2.9158 \end{split}\]
From Equation \((\PageIndex{1.E3.4}\text{b})\)
\[\begin{split} k_{2}^{y} &= f_{1}\left( t_{1} + h,y_{1} + hk_{1}^{y},z_{1} + hk_{1}^{z} \right)\\ &= f_{1}\left( 0.25 + 0.25,1.375 + \left( 0.25 \right)(1.1598),1.1598 + \left( 0.25 \right)\left( - 2.9158 \right) \right)\\ &= f_{1}\left( 0.50,1.6649,0.43087 \right)\\ &= 0.43087 \end{split}\]
From Equation \((\PageIndex{1.E3.6}\text{b})\)
\[\begin{split} k_{2}^{z} &= f_{2}\left( t_{1} + h,y_{1} + hk_{1}^{y},z_{1} + hk_{1}^{z} \right)\\ &= f_{2}\left( 0.25 + 0.25,1.375 + \left( 0.25 \right)(1.1598),1.1598 + \left( 0.25 \right)\left( - 2.9158 \right) \right)\\ &= f_{2}\left( 0.50,1.6649,0.43087 \right)\\ &= e^{- 0.50} - 2\left( 0.43087 \right) - 1.6649\\ &= - 1.9201 \end{split}\]
From Equation \((\PageIndex{1.E3.3})\)
\[\begin{split} y_{2}\ &= y_{1} + \frac{1}{2}\left( k_{1}^{y} + k_{2}^{y} \right)h\\ &= 1.375 + \frac{1}{2}\left( 1.1598 + 0.43087 \right)\left( 0.25 \right)\\ &= 1.5738 \end{split}\]
\(y_{2}\) is the approximate value of \(y\) at
\[t = t_{2} = t_{1} + h = 0.25 + 0.25 = 0.50 \nonumber\]
\[y_{2} = y\left( 0.50 \right) \approx 1.5738 \nonumber\]
From Equation \((\PageIndex{1.E3.5})\)
\[\begin{split} z_{2} &= z_{1} + \frac{1}{2}\left( k_{1}^{z} + k_{2}^{z} \right)h\\ &= 1.1598 + \frac{1}{2}( - 2.9158 + ( - 1.9201))(0.25)\\ &= 0.55533 \end{split}\]
\(z_{2}\) is the approximate value of \(z\) at
\[t = t_{2} = 0.50 \nonumber\]
\[z_{2} = z\left( 0.50 \right) \approx 0.55533 \nonumber\]
For \(i = 2, \ t_{2} = 0.50, \ y_{2} = 1.57384, \ z_{2} = 0.55533\)
From Equation \((\PageIndex{1.E3.4}\text{a})\)
\[\begin{split} k_{1}^{y} &= f_{1}\left( t_{2},y_{2},z_{2} \right)\\ &= f_{1}\left( 0.50,1.5738,0.55533 \right)\\ &= 0.55533 \end{split}\]
From Equation \((\PageIndex{1.E3.6}\text{a})\)
\[\begin{split} k_{1}^{z} &= f_{2}\left( t_{2},y_{2},z_{2} \right)\\ &= f_{2}\left( 0.50,1.5738,0.55533 \right)\\ &= e^{- 0.50} - 2\left( 0.55533 \right) - 1.5738\\ &= - 2.0779 \end{split}\]
From Equation \((\PageIndex{1.E3.4}\text{b})\)
\[\begin{split} k_{2}^{y} &= f_{2}\left( t_{2} + h,y_{2} + hk_{1}^{y},z_{2} + hk_{1}^{z} \right)\\ &= f_{1}\left( 0.50 + 0.25,1.5738 + \left( 0.25 \right)(0.55533),0.55533 + \left( 0.25 \right)\left( - 2.0779 \right) \right)\\ &= f_{1}\left( 0.75,1.7126,0.035836 \right)\\ &= 0.035836 \end{split}\]
From Equation \((\PageIndex{1.E3.6}\text{b})\)
\[\begin{split} k_{2}^{z} &= f_{2}\left( t_{2} + h,y_{2} + hk_{1}^{y},z_{2} + hk_{1}^{z} \right)\\ &= f_{2}\left( 0.50 + 0.25,1.5738 + \left( 0.25 \right)(0.55533),0.55533 + \left( 0.25 \right)\left( - 2.0779 \right) \right)\\ &= f_{2}\left( 0.75,1.7126,0.035836 \right)\\ &= e^{- 0.75} - 2\left( 0.035836 \right) - 1.7126\\ &= - 1.3119 \end{split}\]
From Equation \((\PageIndex{1.E3.3})\)
\[\begin{split} y_{3} &= y_{2} + \frac{1}{2}\left( k_{1}^{y} + k_{2}^{y} \right)h\\ &= 1.5738 + \frac{1}{2}\left( 0.55533 + 0.035836 \right)\left( 0.25 \right)\\ &= 1.6477 \end{split}\]
\(y_{3}\) is the approximate value of \(y\) at
\[t = t_{3} = t_{2} + h = 0.50 + 0.25 = 0.75 \nonumber\]
\[y_{3} = y\left( 0.75 \right) \approx 1.6477 \nonumber\]
b) From Equation \((\PageIndex{1.E3.5})\)
\[\begin{split} z_{3} &= z_{2} + \frac{1}{2}\left( k_{1}^{z} + k_{2}^{z} \right)h\\ &= 0.55533 + \frac{1}{2}( - 2.0779 + ( - 1.3119))(0.25)\\ &= 0.13158 \end{split}\]
\(z_{3}\) is the approximate value of \(z\) at
\[t = t_{3} = 0.75 \nonumber\]
\[z_{3} = z\left( 0.75 \right) \cong 0.13158 \nonumber\]
The intermediate and the final results are shown in Table \(\PageIndex{1.1}\).
| \(i\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|
| \(t_i\) | \(0\) | \(0.25\) | \(0.50\) |
| \(y_i\) | \(1\) | \(1.3750\) | \(1.5738\) |
| \(z_i\) | \(2\) | \(1.1598\) | \(0.55533\) |
| \(k_1^y\) | \(2\) | \(1.1598\) | \(0.55533\) |
| \(k_1^z\) | \(-4\) | \(-2.9158\) | \(-2.0779\) |
| \(k_2^y\) | \(1\) | \(0.43087\) | \(0.035836\) |
| \(k_2^z\) | \(-2.7211\) | \(-1.9201\) | \(-1.3119\) |
| \(y_{i+1}\) | \(1.3750\) | \(1.5738\) | \(1.6477\) |
| \(z_{i+1}\) | \(1.1598\) | \(0.55533\) | \(0.13158\) |
Lesson 2: State-Space Modeling and Higher-Order and Coupled Ordinary Differential Equations
Learning Objectives
After successful completion of this lesson, you should be able to:
1) set up higher-order and coupled ordinary differential equations as simultaneous first-order ordinary differential equations
2) write higher-order and coupled ordinary differential equations in a state-space model form
3) solve higher-order and coupled ordinary differential equations numerically using software programs
Introduction
In the previous lesson, we showed how to write a higher-order ordinary differential equation as a set of first-order differential equations along with the corresponding initial conditions. In this lesson, we show how we rewrite the set of first-order differential equations in matrix form called as the state-space model. We illustrate this through an example. We show the concept through higher-order and coupled ordinary differential equations.
Write the following ordinary differentiation equation as a state model
\[17\frac{d^{3}y}{dt^{3}} + 3\frac{d^{2}y}{dt^{2}} + 7\frac{dy}{dt} + 5y = 11e^{- t}, \nonumber\]
\[y\left( 0 \right) = 13,\ \frac{dy}{dt}\left( 0 \right) = 19,\ \frac{d^{2}y}{dt^{2}} = 23 \nonumber\]
Solution
The differential equation is of the third order, and we have three state variables. Let these be named as
\[x_{1} = y\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E1.1}) \nonumber\]
\[x_{2} = \dot{y}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E1.2}) \nonumber\]
\[x_{3} = \ddot{y}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E1.3}) \nonumber\]
Note the symbols. The symbol \(\dot{y}\) stands for \(\frac{dy}{dt}\) and \(\ddot{y}\) stands for \(\frac{d^{2}y}{dt^{2}}\). We are using these symbols as they are the norm in most textbooks.
We also have
\[\dot{x_{3}} = \dddot{y}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E1.4}) \nonumber\]
So the given ordinary differential equation
\[17\frac{d^{3}y}{dt^{3}} + 3\frac{d^{2}y}{dt^{2}} + 7\frac{dy}{dt} + 5y = 11e^{- t}, \nonumber\]
can be written as
\[17\dot{x_{3}} + 3x_{3} + 7x_{2} + 5x_{1} = 11e^{- t}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E1.5}) \nonumber\]
Writing equations \((\PageIndex{2.E1.2})\), \((\PageIndex{2.E1.3})\), and \((\PageIndex{2.E1.5})\) with the first derivatives on the left side, we get
\[\dot{x_{1}} = \dot{y} = x_{2} \nonumber\]
\[\dot{x_{2}} = \ddot{y} = x_{3} \nonumber\]
\[\dot{x_{3}} = \frac{11e^{- t} - 3x_{3} - 7x_{2} - 5x_{1}}{17}\nonumber\]
Rewriting the above three equations with coefficients for all the state variables, we get
\[\dot{x_{1}} = 0x_{1} + 1x_{2} + 0x_{3} + 0 \nonumber\]
\[\dot{x_{2}} = 0x_{1} + 0x_{2} + 1x_{3} + 0 \nonumber\]
\[\dot{x_{3}} = - \frac{5}{17}x_{1} - \frac{7}{17}x_{2} - \frac{3}{17}x_{3} + \frac{11e^{- t}}{17} \nonumber\]
In the matrix form, the state-space model is given as
\[\begin{bmatrix} \dot{x_{1}} \\ \dot{x_{2}} \\ \dot{x_{3}} \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ - \displaystyle\frac{5}{17} & - \displaystyle\frac{7}{17} & - \displaystyle\frac{3}{17} \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \displaystyle\frac{11e^{- t}}{17} \\ \end{bmatrix} \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E1.6}) \nonumber\]
where the conditions are given by
\[\begin{bmatrix} x_{1}(0) \\ x_{2}(0) \\ x_{3}(0) \\ \end{bmatrix} = \begin{bmatrix} y(0) \\ \dot{y}(0) \\ \ddot{y}(0) \\ \end{bmatrix} = \begin{bmatrix}\displaystyle y(0) \\\displaystyle \frac{dy}{dt}\left( 0 \right) \\\displaystyle \frac{d^{2}y}{dt^{2}}(0) \\ \end{bmatrix} = \begin{bmatrix} 13 \\ 19 \\ 23 \\ \end{bmatrix} \nonumber\]
and
\[y = \begin{bmatrix} 1 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} \nonumber\]
In Equation \((\PageIndex{2.E1.6})\), the \(3 \times 3\) matrix below
\[[A] = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ - \displaystyle\frac{5}{17} & - \displaystyle\frac{7}{17} & - \displaystyle\frac{3}{17} \\ \end{bmatrix} \nonumber\]
is the state matrix.
The 3-element column vector in Equation \((\PageIndex{1.E1.6})\)
\[[X]=\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} \nonumber\]
is called the state vector.
The 3-element column vector in Equation \((\PageIndex{1.E1.6})\)
\[[u] = \begin{bmatrix} 0 \\ 0 \\ \displaystyle\frac{11e^{- t}}{17} \\ \end{bmatrix} \nonumber\]
is called the input vector or forcing function vector. The left side of the equation consists of the first derivatives of the state vector.
The general form of the state-space model is
\[[\dot{X}] = [A][X]+[u] \nonumber\]
Reduce the following coupled ordinary differential equations to a set of first-order differential equations complete with initial conditions and in the matrix form required to solve them numerically.
\[10\frac{d^2x_1}{dt^2} - 15\left(-2x_1 + x_2 \right) = 0 \nonumber\]
\[20 \frac{d^2x_2}{dt^2} - 15\left(x_1-2x_2 \right) = 0 \nonumber\]
\[\frac{dx_1}{dt}\left(0\right) = 4,\ \frac{dx_2}{dt}\left(0\right) = 5,\ x_1\left(0\right) = 2,\ x_2\left(0\right) = 3 \nonumber\]
Solution
The differential equations given are second-order ordinary differential equations
\[10\frac{d^2x_1}{dt^2} - 15\left(-2x_1 + x_2 \right) = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.1}) \nonumber\]
\[20 \frac{d^2x_2}{dt^2} - 15\left(x_1-2x_2 \right) = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.2}) \nonumber\]
The coupled ordinary differential equations \((\PageIndex{2.E2.1})\) and \((\PageIndex{2.E2.2})\) would be rewritten as first-order ordinary differential equations.
Assuming
\[\frac{dx_1}{dt} = x_{3}, \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.3}) \nonumber\]
\[\frac{dx_2}{dt} = x_{4} \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.4}) \nonumber\]
then
\[\frac{d^{2}x_{1}}{dt^2} = \frac{dx_{3}}{dt},\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.5}) \nonumber\]
\[\frac{d^{2}x_{2}}{dt^2} = \frac{dx_{4}}{dt}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.6}) \nonumber\]
Substituting Equations \((\PageIndex{2.E2.3})\), \((\PageIndex{2.E2.4})\), \((\PageIndex{2.E2.5})\), and \((\PageIndex{2.E2.6})\) with the two new variables \(x_{3}\) and \(x_{4}\) in the given second-order ordinary differential equation \((\PageIndex{2.E2.1})\) gives
\[10\frac{dx_{3}}{dt} - 15\left(-2x_{1} + x_{2}\right) = 0 \nonumber\]
and then rewriting it gives
\[\frac{dx_{3}}{dt} = 1.5\left(-2x_1+x_2\right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.7}) \nonumber\]
Substituting equations \((\PageIndex{2.E2.3})\), \((\PageIndex{2.E2.4})\), \((\PageIndex{2.E2.5})\), \((\PageIndex{2.E2.6})\) with the two new variables \(x_{3}\) and \(x_{4}\) in the given second-order ordinary differential equation \((\PageIndex{2.E2.2})\) gives
\[20\frac{dx_4}{dt} - 15\left(x_{1} - 2x_{2}\right) = 0 \nonumber\]
and then rewriting it gives
\[\frac{dx_4}{dt} = 0.75\left(x_{1} - 2x_{2}\right)\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.8}) \nonumber\]
Now, writing the equations \((\PageIndex{2.E2.3})\), \((\PageIndex{2.E2.4})\), \((\PageIndex{2.E2.7})\), and \((\PageIndex{2.E2.8})\) along with the initial conditions \[\frac{dx_{1}}{dt} = x_3,\ x_1(0) = 4,\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.9}\text{a}) \nonumber\]
\[\frac{dx_2}{dt} = x_4,\ x_2(0) = 5,\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.9}\text{b}) \nonumber\]
\[\frac{dx_3}{dt} = 1.5(-2x_1+x_2),\ x_3(0) = 2,\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.9}\text{c}) \nonumber\]
\[\frac{dx_4}{dt} = 0.75(x_1-2x_2),\ x_4(0) = 3,\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.9}\text{d}) \nonumber\]
Preparing for the state-space model form, Equations \((\PageIndex{2.E2.9}\text{a})\) through \((\PageIndex{2.E2.9}\text{d})\) gives
\[\frac{dx_1}{dt} = 0x_1+0x_2+ 1x_3+0x_4,\ x_1(0) = 4,\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.10}\text{a}) \nonumber\]
\[\frac{dx_2}{dt} = 0x_1 + 0x_2+ 0x_3+ 1x_4,\ x_2(0) = 5,\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.10}\text{b}) \nonumber\]
\[\frac{dx_3}{dt} = -3x_1 +1.5x_2 + 0x_3 +0x_4,\ x_3(0) = 2,\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.10}\text{c}) \nonumber\]
\[\frac{dx_4}{dt} = 0.75x_1 -1.5x_2 +0x_3+0x_4,\ x_4(0) = 3,\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E2.10}\text{d}) \nonumber\]
Equations \((\PageIndex{2.E2.10}\text{a})\) through \((\PageIndex{2.E2.10}\text{d})\) can now be written in state-space model form as
\[\left[\begin{array}{l} \dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3} \\ \dot{x}_{4} \end{array}\right]=\left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -3 & 1.5 & 0 & 0 \\ 0.75 & -1.5 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right] \nonumber\]
with the initial conditions as
\[\left[\begin{array}{l} x_{1}(0) \\ x_{2}(0) \\ x_{3}(0) \\ x_{4}(0) \end{array}\right]=\left[\begin{array}{l} 2 \\ 3 \\ 4 \\ 5 \end{array}\right] \nonumber\]
Can you identify the state vector, state matrix, input vector, and the initial conditions on the state vector?
A suspension system of a bus can be modeled as below.
Only 1/4th of the bus is modeled. The differential equations that govern the above system can be derived (this is something you will do in your vibrations course) as
\[M_{1}\frac{d^{2}x_{1}}{dt^{2}} + B_{1}\left( \frac{dx_{1}}{dt} - \frac{dx_{2}}{dt} \right) + K_{1}\left( x_{1} - x_{2} \right) = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.1}) \nonumber\]
\[\begin{split} M_{2}\frac{d^{2}x_{2}}{dt^{2}} + B_{1}\left( \frac{dx_{2}}{dt} - \frac{dx_{1}}{dt} \right) &+ K_{1}\left( x_{2} - x_{1} \right) + B_{2}\left( \frac{dx_{2}}{dt} - \frac{dw}{dt} \right)\\ & \ \ \ \ \ \ + K_{2}\left( x_{2} - w \right) = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.2}) \end{split}\]
\[x_{1}(0) = 0, \ x_{3}(0) = 0, \ x_{2}(0) = 0, \ x_{4}(0) = 0 \nonumber\]
where
\[M_{1} =\text{body mass} \nonumber\]
\[M_{2} =\text{suspension mass} \nonumber\]
\[K_{1} =\text{spring constant of the suspension system} \nonumber\]
\[K_{2} =\text{spring constant of wheel and tire} \nonumber\]
\[B_{1} =\text{damping constant of the suspension system} \nonumber\]
\[B_{2} =\text{damping constant of wheel and tire} \nonumber\]
\[x_{1} =\text{displacement of the body mass as a function of time} \nonumber\]
\[x_{2} =\text{displacement of the suspension mass as a function of time} \nonumber\]
\[w =\text{input profile of the road as a function of time} \nonumber\]
The constants are given as
\[m_{1} = 2500\ \text{kg} \nonumber\]
\[m_{2} = 320\ \text{kg} \nonumber\]
\[k_{1} = 80000\ \text{N/m}, \nonumber\]
\[k_{2} = 500000\ \text{N/m}, \nonumber\]
\[B_{1} = 350\ \text{N-s/m}, \nonumber\]
\[B_{2} = 15020\ \text{N} \cdot \text{s/m} \nonumber\]
Reduce the simultaneous differential equations \((\PageIndex{2.E3.1})\) and \((\PageIndex{2.E3.2})\) to simultaneous first-order differential equations and put those in the state variable form complete with corresponding initial conditions.
Solution
Substituting the values of the constants in the two differential equations \((\PageIndex{2.E3.1})\) and \((\PageIndex{2.E3.2})\) gives the differential equations \((\PageIndex{2.E3.3})\) and \((\PageIndex{2.E3.4})\), respectively.
\[2500\frac{d^{2}x_{1}}{dt^{2}} + 350\left( \frac{dx_{1}}{dt} - \frac{dx_{2}}{dt} \right) + 80000\left( x_{1} - x_{2} \right) = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.3}) \nonumber\]
\[\begin{split} &320\frac{d^{2}x_{2}}{dt^{2}} + 350\left( \frac{dx_{2}}{dt} - \frac{dx_{1}}{dt} \right) + 80000\left( x_{2} - x_{1} \right) + 15020\left( \frac{dx_{2}}{dt} - \frac{dw}{dt} \right) +\\ & 500000(x_{2} - w) = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.4}) \end{split}\]
Since \(w\) is an input, we take it to the right-hand side to show it as a forcing function and rewrite Equations \((\PageIndex{2.E3.4})\) as
\[\begin{split} &320\frac{d^{2}x_{2}}{dt^{2}} + 350\left( \frac{dx_{2}}{dt} - \frac{dx_{1}}{dt} \right) + 80000\left( x_{2} - x_{1} \right) + 15020\left( \frac{dx_{2}}{dt} \right) + 500000x_{2}\\ &=15020\frac{dw}{dt}+500000w\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.5}) \end{split}\]
Now let us start the process of reducing the 2 simultaneous differential equations (Equations \((\PageIndex{2.E3.3})\) and \((\PageIndex{2.E3.5})\)) to 4 simultaneous first-order ordinary differential equations.
Choose
\[\frac{dx_{1}}{dt} = x_{3},\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.6}) \nonumber\]
\[\frac{dx_{2}}{dt} = x_{4},\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.7}) \nonumber\]
then Equation \((\PageIndex{2.E3.3})\)
\[2500\frac{d^{2}x_{1}}{dt^{2}} + 350\left( \frac{dx_{1}}{dt} - \frac{dx_{2}}{dt} \right) + 80000\left( x_{1} - x_{2} \right) = 0 \nonumber\]
can be written as
\[2500\frac{dx_{3}}{dt} + 350\left( x_{3} - x_{4} \right) + 80000\left( x_{1} - x_{2} \right) = 0 \nonumber\]
\[2500\frac{dx_{3}}{dt} = - 350\left(x_{3} - x_{4}\right) - 80000\left(x_{1} - x_{2}\right) \nonumber\]
\[\begin{split} \frac{dx_{3}}{dt} &= - 0.14\left( x_{3} - x_{4} \right) - 32\left( x_{1} - x_{2} \right)\\ &= - 32x_{1} - 0.14x_{3} + 32x_{2} + 0.14x_{4}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.8})\end{split}\]
and Equation \((\PageIndex{2.E3.5})\)
\[\begin{split} &320\frac{d^{2}x_{2}}{dt^{2}} + 350\left( \frac{dx_{2}}{dt} - \frac{dx_{1}}{dt} \right) + 80000\left( x_{2} - x_{1} \right) + 15020\left( \frac{dx_{2}}{dt} \right) + 500000x_{2} \\ &= 15020\frac{dw}{dt} + 500000w \end{split}\]
can be written as
\[\begin{split} &320\frac{dx_{4}}{dt} + 350\left( x_{4} - x_{3} \right) + 80000\left( x_{2} - x_{1} \right) + 15020x_{4} + 500000x_{2}\\ &= 15020\frac{dw}{dt} + 500000w \end{split}\]
\[\begin{split} 320\frac{dx_{4}}{dt} = &- 350(x_{4} - x_{3}) - 80000(x_{2} - x_{1}) - 15020x_{4} - \\ & 500000x_{2} + 15020\frac{dw}{dt} + 500000w \end{split}\]
\[\begin{split} \frac{dx_{4}}{dt} &= - 1.09375\left( x_{4} - x_{3} \right) - 250\left( x_{2} - x_{1} \right) - 46.9375x_{4} - 1562.5x_{2}\\& \ \ \ \ \ \ \ \ \ \ + 46.9375\frac{dw}{dt} + 1562.5w\\ &= 250x_{1} + 1.09375x_{3} - 1812.5x_{2} - 48.03125x_{4}\\ &\ \ \ \ \ \ \ \ \ \ + 1562.5w + 46.9375\frac{dw}{dt}\ \ \ \ \ \ \ \ \ (\PageIndex{2.E3.9}) \end{split}\]
The 4 simultaneous first-order differential equations given by Equations \((\PageIndex{2.E3.6})\) through \((\PageIndex{2.E3.9})\), complete with the corresponding initial condition, are
\[\begin{split} \frac{dx_{1}}{dt} &= x_{3}\\ &=f_1\left(t,x_1,x_2,x_3,x_4\right)\ \text{with } x_1\left(0\right)=0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.10}) \end{split}\]
\[\begin{split} \frac{dx_{2}}{dt} &= x_{4}\\ &=f_2\left(t,x_1,x_2,x_3,x_4\right)\text{with }x_2\left(0\right)=0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.11}) \end{split}\]
\[\begin{split} \frac{dx_{3}}{dt} &= - 32x_{1} + 32x_{2} - 0.14x_{3} + 0.14x_{4}\\ &=f_3\left(t,x_1,x_2,x_3,x_4\right)\text{with }x_3\left(0\right)=0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.12}) \end{split}\]
\[\begin{split} \frac{dx_{4}}{dt} &= 250x_{1} - 1812.5x_{2} + 1.09375x_{3} - 48.03125x_{4} + 1562.5w + 46.9375\frac{dw}{dt}\\ &=f_4\left(t,x_1,x_2,x_3,x_4\right)\text{with }x_4\left(0\right)=0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.13}) \end{split}\]
The profile of the road is given below.
Assuming that the bus is going at \(60 \ \text{mph}\), that is, approximately \(27 \ \text{m/s}\), it takes
\[\frac{6 \ \text{m}}{27 \ \text{m/s}} = 0.22\ \text{s} \nonumber\]
to go through one period. So the frequency
\[\begin{split} f &= \frac{1}{0.22}\\ &= 4.545\ \text{Hz}\end{split}\]
The angular frequency then is
\[\begin{split} \omega &=2 \times \pi \times 4.545\\ &= 28.6\ \text{rad/s} \end{split}\]
giving
\[\begin{split} w &= 0.01\sin\left( {\omega t} \right)\\ &= 0.01\sin\left( 28.6t \right)\end{split}\]
and
\[\frac{dw}{dt} = 0.286\cos\left( 28.6t \right) \nonumber\]
To put the differential equations given by Equations \((\PageIndex{2.E3.10})\) through \((\PageIndex{2.E3.13})\) in matrix form, we rewrite them as
\[\frac{dx_{1}}{dt} = x_{3} = 0x_{1} + 0x_{2} + 1x_{3} + 0x_{4},\ {x}_{1}\left( 0 \right) = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.14}) \nonumber\]
\[\frac{dx_{2}}{dt} = x_{4} = 0x_{1} + 0x_{2} + 0x_{3} + 1x_{4},\ {x}_{2}\left( 0 \right) = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.15}) \nonumber\]
\[\frac{dx_{3}}{dt} = - 32x_{1} + 32x_{2} - 0.14x_{3} + 0.14x_{4},\ x_{3}\left( 0 \right) = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.16}) \nonumber\]
\[\begin{split} \frac{dx_{4}}{dt} = &250x_{1} - 1812.5x_{2} + 1.09375x_{3} - 48.03125x_{4} + \\ &1562.5w + 46.9375\frac{dw}{dt},\ x_{4}\left( 0 \right) = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.17}) \end{split}\]
In state variable matrix form, the differential equations are given by
\[\begin{split} \begin{bmatrix} \displaystyle \frac{dx_{1}}{dt} \\ \displaystyle \frac{dx_{2}}{dt} \\ \displaystyle \frac{dx_{3}}{dt} \\ \displaystyle \frac{dx_{4}}{dt} \\ \end{bmatrix} &= \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 32 & 32 & - 0.14 & 0.14 \\ 250 & - 1812.5 & 1.09375 & - 48.03125 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ \end{bmatrix} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1562.5w + 46.9375\displaystyle \frac{dw}{dt} \\ \end{bmatrix} \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.18})\end{split}\]
where
\[w = 0.01\sin(28.6t) \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.E3.19}) \nonumber\]
and the corresponding initial conditions are
\[\begin{bmatrix} x_{1}(0) \\ x_{2}(0) \\ x_{3}(0) \\ x_{4}(0) \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} \nonumber\]
Can you identify the state vector, the state matrix, the input vector, and the initial conditions on the state vector?
State-space models is how numerical solvers expect you to enter the data. For example, the above state-space model would be entered as the following on MATLAB.
First, a function is defined where t is the independent variable, and x is the state-space vector. The names diffx and sysofeqn are the choice of the user. The arguments of the function are the independent variable t and the state vector x. Inside the function, \([A]\) is the state-space matrix from Equation \((\PageIndex{2.E3.18})\), \(w\) is the forcing function from Equation \((\PageIndex{2.E3.19})\), and \([B]\) is the input vector.
function diffx=sysofeqn(t,x)
A=[0 0 1 0; …
0 0 0 1; …
-32 32 -0.14 0.14; …
250 -1812.5 1.09375 -48.03125];
w=0.01*sin(28.6*t);
dw=0.286*cos(28.6*t);
B=[0; 0; 0; 1652.5*w+46.9375*dw];
diffx=A*x+B;
end
How is the function used? Assume that you want the output between \(t=0\) and \(t=10\). You can use that time span shown as the variable tspan. Then the initial conditions of the state vector are entered. The ode45 routine is a combination of the Runge-Kutta 4th and 5th order methods to solve a system of first-order ordinary differential equations. The inputs of the ode45 routine are the system of equations, the span in which the output is requested, and the initial conditions.
tspan=[0 10];
initial_cond=[0;0;0;0];
[t,x]=ode45('sysofeqn',tspan,initial_cond);
figure (1)
plot(t,x(:,1))
A sample of the outputs is shown below with the values of the independent variable \(t\), the state vector \(x\), and the output of \(x(1)\) vs \(t\).
\[\begin{split} t &= \\ &\begin{array}{c} 0 \\ 0.000003742346606 \\ 0.000007484693212 \\ 0.000011227039818 \\ 0.000014969386423 \\ 0.000033681119453 \\ 0.000052392852482 \end{array} \end{split}\]
\[\begin{split} x &= \\ &\begin{array}{c} 0 \\ 0.000000000000000 \\ 0.000000000000000 \\ 0.000000000000000 \\ 0.000000000000001 \\ 0.000000000000012 \\ 0.000000000000045 \end{array} \quad \begin{array}{c} 0 \\ 0.000000000094002 \\ 0.000000000376002 \\ 0.000000000845991 \\ 0.000000001503960 \\ 0.000000007613185 \\ 0.000000018420550 \end{array} \quad \begin{array}{c} 0 \\ 0.000000000013164 \\ 0.000000000052670 \\ 0.000000000118540 \\ 0.000000000210794 \\ 0.000000001068580 \\ 0.000000002589166 \end{array} \quad \begin{array}{c} 0 \\ 0.000050236522812 \\ 0.000100470633045 \\ 0.000150702329280 \\ 0.000200931610101 \\ 0.000452041733437 \\ 0.000703091259071 \end{array} \end{split}\]
Multiple Choice Test
(1). To solve, the ordinary differential equation
\[2\frac{d^{2}y}{dx^{2}} + 5\frac{dy}{dx} + 6y = 11e^{- x}, \ y(0) = 5, \ \frac{dy}{dx}(0) = 7 \nonumber\]
can be reduced to a net of how many first order differential equations?
(A) 1
(B) 2
(C) 3
(D) 4
(2). Find the state variable form of the following linear dynamic system
\[2\frac{d^{2}x_{1}}{dt^{2}} + 3x_{1} - 5\left( x_{2} - x_{1} \right) - 7\left( \frac{dx_{2}}{dt} - \frac{dx_{1}}{dt} \right) = 0 \nonumber\]
\[11\frac{d^{2}x_{2}}{dt^{2}} + 5\left( x_{2} - x_{1} \right) + 7\left( \frac{dx_{2}}{dt} - \frac{dx_{1}}{dt} \right) = \sin(2t) \nonumber\]
\[\frac{dx_{1}}{dt}(0) = 13,\ \ \frac{dx_{2}}{dt}(0) = 17,\ \ x_{1}(0) = 23,\ \ x_{2}(0) = 29\nonumber\]
where \(x_{1}\) and \(x_{2}\) are dependent variables and one can choose \(v_{1} = \dfrac{dx_{1}}{dt}\) and \(v_{2} = \dfrac{dx_{2}}{dt}\) as newly introduced dependent variables for the state variable form.
(A) \(\ \ \ \ \begin{bmatrix} \dot{x_{1}} \\ \dot{v_{1}} \\ \begin{matrix} \dot{x_{2}} \\ \dot{v_{2}} \\ \end{matrix} \\ \end{bmatrix} = \left\lbrack \begin{matrix} 0 & - 1 & 0 \\ 4 & \frac{7}{2} & - \frac{5}{2} \\ \begin{matrix} 0 \\ - \frac{5}{11} \\ \end{matrix} & \begin{matrix} 0 \\ - \frac{7}{11} \\ \end{matrix} & \begin{matrix} 0 \\ \frac{5}{11} \\ \end{matrix} \\ \end{matrix}\ \ \ \ \begin{matrix} 0 \\ - \frac{7}{2} \\ \begin{matrix} - 1 \\ \frac{7}{11} \\ \end{matrix} \\ \end{matrix} \right\rbrack\begin{bmatrix} x_{1} \\ v_{1} \\ \begin{matrix} x_{2} \\ v_{2} \\ \end{matrix} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \begin{matrix} 0 \\ \sin(2t) \\ \end{matrix} \\ \end{bmatrix}\)
with \(\begin{bmatrix} x_{1}(0) \\ v_{1}(0) \\ \begin{matrix} x_{2}(0) \\ v_{2}(0) \\ \end{matrix} \\ \end{bmatrix} = \begin{bmatrix} 23 \\ 13 \\ \begin{matrix} 29 \\ 17 \\ \end{matrix} \\ \end{bmatrix}\)
(B) \(\ \ \ \ \begin{bmatrix} \dot{x_{1}} \\ \dot{v_{1}} \\ \begin{matrix} \dot{x_{2}} \\ \dot{v_{2}} \\ \end{matrix} \\ \end{bmatrix} = \left\lbrack \begin{matrix} 0 & 1 & 0 \\ - 4 & - \frac{7}{2} & \frac{5}{2} \\ \begin{matrix} 0 \\ \frac{5}{11} \\ \end{matrix} & \begin{matrix} 0 \\ \frac{7}{11} \\ \end{matrix} & \begin{matrix} 0 \\ - \frac{5}{11} \\ \end{matrix} \\ \end{matrix}\ \ \ \ \begin{matrix} 0 \\ \frac{7}{2} \\ \begin{matrix} 1 \\ - \frac{7}{11} \\ \end{matrix} \\ \end{matrix} \right\rbrack\begin{bmatrix} x_{1} \\ v_{1} \\ \begin{matrix} x_{2} \\ v_{2} \\ \end{matrix} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \begin{matrix} 0 \\ \sin(2t) \\ \end{matrix} \\ \end{bmatrix}\)
with \(\begin{bmatrix} x_{1}(0) \\ v_{1}(0) \\ \begin{matrix} x_{2}(0) \\ v_{2}(0) \\ \end{matrix} \\ \end{bmatrix} = \begin{bmatrix} 23 \\ 13 \\ \begin{matrix} 29 \\ 17 \\ \end{matrix} \\ \end{bmatrix}\)
(C) \(\ \ \ \ \begin{bmatrix} \dot{x_{1}} \\ \dot{v_{1}} \\ \begin{matrix} \dot{x_{2}} \\ \dot{v_{2}} \\ \end{matrix} \\ \end{bmatrix} = \left\lbrack \begin{matrix} 0 & 1 & 0 \\ - 4 & - \frac{7}{2} & \frac{5}{2} \\ \begin{matrix} 0 \\ \frac{5}{11} \\ \end{matrix} & \begin{matrix} 0 \\ \frac{7}{11} \\ \end{matrix} & \begin{matrix} 0 \\ - \frac{5}{11} \\ \end{matrix} \\ \end{matrix}\ \ \ \ \begin{matrix} 0 \\ \frac{7}{2} \\ \begin{matrix} 1 \\ - \frac{7}{11} \\ \end{matrix} \\ \end{matrix} \right\rbrack\begin{bmatrix} x_{1} \\ v_{1} \\ \begin{matrix} x_{2} \\ v_{2} \\ \end{matrix} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \begin{matrix} 0 \\ \frac{\sin(2t)}{11} \\ \end{matrix} \\ \end{bmatrix}\)
with \(\begin{bmatrix} x_{1}(0) \\ v_{1}(0) \\ \begin{matrix} x_{2}(0) \\ v_{2}(0) \\ \end{matrix} \\ \end{bmatrix} = \begin{bmatrix} 23 \\ 13 \\ \begin{matrix} 29 \\ 17 \\ \end{matrix} \\ \end{bmatrix}\)
(D) \(\begin{bmatrix} \dot{v_{1}} \\ \dot{v_{2}} \\ \end{bmatrix} = \left\lbrack \begin{matrix} - 4 & - \frac{7}{2} & \frac{5}{2} \\ \frac{5}{11} & \frac{7}{11} & - \frac{5}{11} \\ \end{matrix}\ \ \ \ \begin{matrix} \frac{7}{2} \\ - \frac{7}{11} \\ \end{matrix} \right\rbrack\begin{bmatrix} v_{1} \\ v_{2} \\ \end{bmatrix} + \begin{bmatrix} 0 \\ \frac{\sin(2t)}{11} \\ \end{bmatrix}\)
with \(\begin{bmatrix} v_{1} \\ v_{2} \\ \end{bmatrix} = \begin{bmatrix} 13 \\ 17 \\ \end{bmatrix}\)
(3). Given
\[2\frac{d^{2}y}{dx^{2}} + 5\frac{dy}{dx} + 7y = 6x^{2}, \ y(0) = 9, \ \frac{dy}{dx}(0) = 11 \nonumber,\]
what is the estimated value of \(y(4)\) using Euler’s method? Step size used is \(h=2\).
(A) \(-205.0\)
(B) \(192.0\)
(C) \(9.181\)
(D) \(31\)
(4). Given
\[2\frac{d^{2}y}{dx^{2}} + 5\frac{dy}{dx} + 7y = 6x^{2}, \ y(0) = 9, \ \frac{dy}{dx}(0) = 11 \nonumber,\]
what is the estimated true error of \(y(4)\) using Heun’s method? Step size used is \(h=2\).
(A) \(-205.0\)
(B) \(192.0\)
(C) \(9.181\)
(D) \(-77\)
(5). The differential equation
\[2\frac{d^{2}y}{dx^{2}} + 5\frac{dy}{dx} + 7y = 6x^{2}, \ y(0) = 9, \ \frac{dy}{dx}(0) = 11 \nonumber\]
is solved by using Euler’s method. A step size of \(h=2\) is used and the following output variables are tabled.
| \(X\) | \(Y\) | \(Z\) |
|---|---|---|
| \(0\) | \(9\) | \(11\) |
| \(2\) | \(31\) | \(-107\) |
| \(4\) | \(-205\) | \(235\) |
What is the best estimate of \(\dfrac{dy}{dx}(4)\) and \(\dfrac{d^{2}y}{dx^{2}}(4)\)?
(A) \(\dfrac{dy}{dx}(4) = 235\ ;\ \dfrac{d^{2}y}{dx^{2}}(4) = 178\)
(B) \(\dfrac{dy}{dx}(4) = - 122.5\ ;\ \dfrac{d^{2}y}{dx^{2}}(4) = 178\)
(C) \(\dfrac{dy}{dx}(4) = 235\ ;\ \dfrac{d^{2}y}{dx^{2}}(4) = 171\)
(D) \(\dfrac{dy}{dx}(4) = - 122.5\ ;\ \dfrac{d^{2}y}{dx^{2}}(4) = 171\)
(6). Given the state variable form of a second order ordinary differential equation as
\(\begin{bmatrix} \dot{y} \\ \dot{z} \\ \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ - 3 & - 6 \\ \end{bmatrix}\begin{bmatrix} y \\ z \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 5x^{2} \\ \end{bmatrix}\)
where \(\begin{bmatrix} y(0) \\ z(0) \\ \end{bmatrix} = \begin{bmatrix} 9 \\ 11 \\ \end{bmatrix}\),
then the second order ODE in question is
(A) \(\dfrac{d^{2}y}{dx^{2}} - 6\dfrac{dy}{dx} - 3y = 5x^{2}, \ y(0) = 9, \ \dfrac{dy}{dx}(0) = 11\)
(B) \(\dfrac{d^{2}y}{dx^{2}} + 6\dfrac{dy}{dx} + 3y = 5x^{2}, \ y(0) = 9, \ \dfrac{dy}{dx}(0) = 11\)
(C) \(\dfrac{d^{2}y}{dx^{2}} + 6\dfrac{dy}{dx} + 3y = 5x^{2}, \ y(0) = 11, \ \dfrac{dy}{dx}(0) = 9\)
(D) \(\dfrac{d^{2}y}{dx^{2}} + 6\dfrac{dy}{dx} - 3y = 5x^{2}, \ y(0) = 9, \ \dfrac{dy}{dx}(0) = 11\)
Problem Set
(1). Reduce the following 2nd order ordinary differential equation to a set of first-order ordinary differential equations complete with initial conditions and in the form required to solve them numerically.
\[5\frac{d^{2}y}{dt^{2}} + 3\frac{dy}{dt} + 7y = e^{- t} + t^{2},\ y(0) = 6,\ \frac{dy}{dt}(0) = 11. \nonumber\]
(2). Reduce the following coupled ordinary differential equations to a set of first-order ordinary differential equations complete with initial conditions and in the form required for solving them numerically.
\[10\frac{d^{2}x_{1}}{dt^{2}} - 15( - 2x_{1} + x_{2}) = 0, \nonumber\]
\[20\frac{d^{2}x_{2}}{dt^{2}} - 15(x_{1} - 2x_{2}) = 0, \nonumber\]
\[x_{1}(0) = 2,\ x_{2}(0) = 3,\ \frac{dx_{1}}{dt}(0) = 4,\ \frac{dx_{2}}{dt}(0) = 5. \nonumber\]
(3). The acceleration of a spring-mass system is given by
\[\frac{d^{2}x}{dt^{2}} + x = e^{- t}, \ t > 0 \nonumber\]
where \(x\) is the displacement of the mass given in meters and \(t\) is the time given in seconds. The initial conditions are given as \(x(0) = 3\), \(\displaystyle \frac{dx}{dt}(0) = 2\).
a) What is the value of displacement, velocity, and acceleration at \(t = 0^{+}\) seconds?
b) Use Euler’s method to find the value of displacement, velocity, and acceleration at \(t = 0.5\) seconds. Use a step size of \(0.25\) seconds.
c) What is the exact value of displacement, velocity and acceleration at \(t = 0.5\) seconds.
- Answer
-
\(a)\ 3,\ 2 ,\ -2\)
\(b)\ 3.8750,\ 0.81970,\ -3.2685\)
\(c)\ 3.6958,\ 0.69213,\ -3.0893\)
(4). The acceleration of a spring-mass system is given by
\[\frac{d^{2}x}{dt^{2}}\ + x = \ e^{- t} \nonumber\]
where \(x\) is the displacement of the mass given in meters and \(t\) is the time given in seconds. The initial conditions are given as \(x(0) = 3\), \(\displaystyle \frac{dx}{dt}(0) = 2\). Use the Runge-Kutta 2nd order Ralston’s method to find the value of displacement, velocity, and acceleration at \(t = 0.5\) seconds. Use a step size of \(0.25\) seconds.
- Answer
-
\(3.7067\ \text{m},\ 0.67811\ \text{m/s},\ -3.1001\ \text{m/s}^2\)
(5). Reduce the following ordinary differential equation to the state-space model form. Include the initial conditions in proper form as well.
\[4\frac{d^{2}y}{dt^{2}} + 9\frac{dy}{dt} + 12y = 3 \sin(7x),\ y(0) = 13,\ \frac{dy}{dt}(0) = 19. \nonumber\]
(6). Reduce the following coupled ordinary differential equations to the state space model form. Include the initial conditions in proper form as well.
\[25\frac{d^{2}x_{1}}{dt^{2}} - 40( - 3x_{1} + x_{2}) = 3 \cos(3t), \nonumber\]
\[32\frac{d^{2}x_{2}}{dt^{2}} - 12x_{1} - 24x_{2} = 7e^{-3t}, \nonumber\]
\[x_{1}(0) = 23,\ x_{2}(0) = 17,\ \frac{dx_{1}}{dt}(0) = 37,\ \frac{dx_{2}}{dt}(0) = 29. \nonumber\]