odes mat lab

8/10/2019 Odes Mat Lab

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Using Matlab for First Order ODEs

Contents

@-functionsDirection fieldsNumerical solution of initial value problems

Plotting the solutionCombining direction field and solution curvesFinding numerical values at given t values

Symbolic solution of ODEsFinding the general solutionSolving initial value problems

Plotting the solutionFinding numerical values at given t values

Symbolic solutions Dealing !ith solutions in implicit form

@-functions

"ou can define a function in #atlab using the @-synta$

g % @&$' sin&$'($

defines the function g&$' % sin&$')$* "ou can then

evaluate the functionfor a given $-valueg&+*,'

plot the graph of the functionover an intervalezplot&g.+/+0'

find a zero of the functionnear an initial guessfzero&g,'

"ou can also define @-functions of several variables

1 % @(x,y)$23 4 y23 - 3(&$2/4y2/' 4 3

defines the function 1&$y' % $34 y3- 3&$/4 y/' 4 3 of t!o variables* "ou canthen

evaluate the functionfor given values of $y1&5/'

plot the graph of the functionas a surface over a rectangle in the $yplane

ezsurf&1.-//-//0'

Clic6 on in the figure toolbar then you can rotate the graph bydragging !ith the mouse*
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plot the curves where G(x,y)=0in a rectangle in the $y planeezplot&1.-//-//0'

make a contour plot of the functionfor a rectangle in the $y planeezcontour&1.-//-//0'7 colorbar

Direction FieldsFirst do!nload the file dirfield*mand put it in the same directory as your otherm-files for the home!or6*

Define an @-function f of t!o variables t y corresponding to the right handside of the differential e8uation y9&t' % f&ty&t''* E*g* for the differential

e8uation y9&t' % t y/define

f % @&ty' t(y2/

"ou must use @&ty'*** even if t or y does not occur in your formula*

E*g* for the ODE y9%y/you !ould use f%@&ty'y2/

:o plot the direction field for t going from t+ to t5 !ith a spacing of dt and ygoing from y+ to y5 !ith a spacing of dy use dirfield&ft+dtt5y+dyy5'* E*g*for t and y bet!een -/ and / !ith a spacing of +*/ type

dirfield&f-/+*//-/+*//'

Solving an initial value problem numerically

First define the @-function f corresponding to the right hand side of thedifferential e8uation y9&t' % f&ty&t''* E*g* for the differential e8uation y9&t'

% t y/define

f % @&ty' t(y2/

:o plot the numerical solution of an initial value problem For the initialcondition y&t+'%y+ you can plot the solution for t going from t+ to t5
http://terpconnect.umd.edu/~petersd/246/dirfield.mhttp://terpconnect.umd.edu/~petersd/246/dirfield.m


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using ode3;&f.t+t50y+'*

E$ample :o plot the solution of the initial value problem y9&t' % t y/ y&-/'%5 inthe interval .-//0 use

.tsys0 % ode3;&f.-//05'

plot&tsys9o-9'

:he circles mar6 the values !hich !ere actually computed &the points arechosen by #atlab to optimi


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:o obtain numerical values of the solution at certain t values "ou can specifya vector tv of t values and use .tsys0 % ode3;&gtvy+'* :he first element of the

vector tv is the initial t value7 the vector tv must have at least , elements* E*g*to obtain the solution !ith the initial condition y&-/'%5 at t % -/ -5*; *** 5*; /and display the results as a table !ith t!o columns use

.tsys0%ode3;&f-/+*;/5'7

.tsys0

:o obtain the numerical value of the solution at the final t-value use ys&end' *

"t may happen that the solution does not exist on the whole interval#

f % @&ty' t(y2/.tsys0 % ode3;&f.+/0/'7

?n this case ode45prints a !arning Failure at t=... to sho! !here it

stopped*

Note that in some cases ode$%sperforms better than ode3;* :his happensfor so-called stiff problems* ode5;s is also better at detecting !here asolution stops to e$ist if the slope becomes infinite*

Solving a differential equation symbolically"ou have to specify the differential e8uation in a string using Dy for y9&t'

and y for y&t' E*g* for the differential e8uation y9&t' % t y/type

sol % dsolve&9Dy%t(y2/99t9'

:he last argument 9t9 is the name of the independent variable* Do nottype y&t' instead of y*

?f #atlab can9t find a solution it !ill return an empty symbol* ?f #atlab finds

several solutions it returns a vector of solutions*

Aere there are t!o solutions and #atlab returns a vector sol!ith t!o


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components sol(1)is 0and sol(2)is -1/(t^2/2 + C3)!ith an arbitrary

constant C3*

:he solution !ill contain a constant C, &or C3C; etc*'* "ou can substitutevalues for the constant using subs&sol9C,9value'* E*g* to set C, in sol&/' to ;use

subs&sol&/'9C,9;'

:o solve an initial value problem additionally specify an initial condition

sol % dsolve&9Dy%t(y2/99y&-/'%599t9'

:o plot the solution use e


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ootOf&G>H2; - ;(G>H - &;(t2/'/ 4 ;/ G>H':his means that the solution in implicit form is

y;- ;y - ;t// 4 ;/ % +2.dsolve returns the ans!er empty sym !?n this case #atlab !as unable to find the solution in implicit form* ?n

older versions &e*g* #atlab /+5+b' this can even happen !hen it easyto find by hand the solution in implicit form* ?n some cases omitting theinitial condition helpsFor E$ample 5 ne!er #atlab versions &/+55b /+5/b' return .emptysym0* ?n this case using dsolve&9Dy%t&y23-5'99t9' gives the implicitsolution !ith a constant* Ie can then find the value of the constantusing the initial condition*

lotting the solution of "* in implicit form#?f the solution in implicit formis expression%+ use

ezplot(expression,[ tmintmaxyminymin])

to plot the solution y&t' for tmin t J t ma$ ymin y J y ma$*E*g* for E$ample 5 !e can plot the initial point together !ith the solutioncurve by

$old o!% plot(1&0&'o')%

eplot('^5 - 5* + 5/2 - 5*t^2/2'&-2 2 -2 2,)% rid o!% $old

o##

Ie see from the graph that the interval where the solution existsis roughly&-5*K 5*K'*

lotting the general solution in implicit form#?f the general solution inimplicit form is expression%C !ith C arbitrary use

ezcontour(expression, tmin

tmax

ymin

ymin

!)

E*g* for E$ample 5 !e can plot the general solution bye


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t%5*;7 fzero&@&y'y2;-;(y4;/-;(t2//-+*;'!hich returns the ans!er y%-+*K3L>5H*

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