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1. Matlab R2012b With Key
2. Matlab R2012b With Usb
3. Matlab R2012b With Crack

Contents

@-functions

Direction fields

Numerical solution of initial value problems

Plotting the solution
Combining direction field and solution curves
Finding numerical values at given t values

Symbolic solution of ODEs

Finding the general solution
Solving initial value problems
Plotting the solution
Finding numerical values at given t values

Symbolic solutions: Dealing with solutions in implicit form

@-functions

You can define a function in Matlab using the @-syntax:

g = @(x) sin(x)*x

defines the function g(x) = sin(x)·x. You can then

• evaluate the function for a given x-value:
  g(0.3)
• plot the graph of the function over an interval:
  ezplot(g,[0,20])
• find a zero of the function near an initial guess:
  fzero(g,3)

You can also define @-functions of several variables:

G = @(x,y) x^4 + y^4 - 4*(x^2+y^2) + 4

defines the function G(x,y) = x⁴ + y⁴ - 4(x² + y²) + 4 of two variables.You can then

• evaluate the function for given values of x,y:
  G(1,2)
• plot the graph of the function as a surface over a rectangle in the x,y plane:
  ezsurf(G,[-2,2,-2,2])
  Click on in the figure toolbar, then you can rotate the graph by dragging with the mouse.
• plot the curves where G(x,y)=0 in a rectangle in the x,y plane:
  ezplot(G,[-2,2,-2,2])
• make a contour plot of the function for a rectangle in the x,y plane:
  ezcontour(G,[-2,2,-2,2]); colorbar

Direction Fields

First download the file dirfield.m and put it inthe same directory as your other m-files for the homework.

Define an @-function fof two variables t, y corresponding to the righthand side of the differential equation y'(t) =f(t,y(t)). E.g., for the differentialequation y'(t) = t y²define

f = @(t,y) t*y^2

You must use@(t,y)..., even ift or y does not occur in your formula.
E.g., for the ODE y'=y² you would use f=@(t,y)y^2

To plot the direction field for t going from t0 to t1 witha spacing of dt and y going from y0 to y1 with a spacing of dy usedirfield(f,t0:dt:t1,y0:dy:y1). E.g., fort and y between -2 and 2 with a spacing of 0.2type

dirfield(f,-2:0.2:2,-2:0.2:2)

Solving an initial value problem numerically

First define the @-functionfcorresponding to theright hand side of the differential equation y'(t) =f(t,y(t)). E.g., for the differentialequation y'(t) = t y²define

f = @(t,y) t*y^2

To plot the numerical solution of an initial valueproblem: For the initial condition y(t0)=y0 you can plot thesolution for t going from t0 to t1 usingode45(f,[t0,t1],y0).

Example: To plot the solution of the initial value problemy'(t) = t y², y(-2)=1in the interval [-2,2] use

[ts,ys] = ode45(f,[-2,2],1)
plot(ts,ys,'o-')

The circles mark the values which were actually computed (the points arechosen by Matlab to optimize accuracy and efficiency). The vectorsts and ys contain the coordinates of these points, to see them as a table type [ts,ys]

You can plotthe solution without the circles using plot(ts,ys).

To combine plots of the direction field and severalsolution curves use the commands holdon and hold off: After obtainingthe first plot type hold on, then all subsequent commands plot inthe same window. After the last plot command type hold off.

Example: Plot the direction field and the 13 solution curves withthe initial conditions y(-2) = -0.4, -0.2, ..., 1.8, 2:

To obtain numerical values of the solution atcertain t values: You can specify a vector tv of tvalues and use [ts,ys] = ode45(g,tv,y0). Thefirst element of the vector tv is the initial t value; the vectortv must have at least 3 elements. E.g., to obtain the solutionwith the initial condition y(-2)=1 at t = -2, -1.5, ..., 1.5, 2 anddisplay the results as a table with two columns, use

[ts,ys]=ode45(f,-2:0.5:2,1);
[ts,ys]

To obtain the numerical value of the solution at the finalt-value useys(end) .

It may happen that the solution does not exist on the whole interval:

f = @(t,y) t*y^2
[ts,ys] = ode45(f,[0,2],2);

In this case ode45 prints a warning 'Failure at t=...' to show where it stopped.

Note that in some cases ode15s performs better than ode45.This happens for so-called stiff problems. ode15s is also better at detecting wherea solution stops to exist if the slope becomes infinite.

Solving a differential equation symbolically

sol = dsolve('Dy=t*y^2','t')

The last argument 't' is the name of the independent variable.Do not type y(t) instead of y.

If Matlab can't find a solution it will return an empty symbol. If Matlabfinds several solutions it returns a vector of solutions.

Here there are two solutions and Matlab returns a vector sol with two components:sol(1) is 0 and sol(2) is -1/(t^2/2 + C3)with an arbitrary constant C3.

The solution will contain a constant C3 (or C4,C5 etc.). You can substitutevalues for the constant usingsubs(sol,'C3',value). E.g., to setC3 in sol(2) to 5 use

subs(sol(2),'C3',5)

To solve an initial value problemadditionally specify an initial condition:

sol = dsolve('Dy=t*y^2','y(-2)=1','t')

To plot the solution useezplot(sol,[t0,t1]). Here is an example forplotting the solution curve with the initial conditions y(-2) =-0.4:

To obtain numerical values at one ormore t values use subs(sol,'t',tval) anddouble (or vpa formore digits):

sol = dsolve('Dy=t*y^2','y(-2)=1','t')

This gives a numerical value of the solution at t=0.5:

double( subs(sol,'t',0.5) )

This computes numerical values of the solution at t=-2, -1.5, ..., 2 anddisplays the result as a table with two columns:

tval = (-2:0.5:2)'; % column vector with t-values
yval = double( subs(sol,'t',tval) )% column vector with y-values
[tval,yval] % display 2 columns together

Symbolic solutions: Dealing with solutions in implicit form

Matlab R2012b With Key

Often dsolve says 'Explicit solution could not be found'.But in many cases one can still obtain the solution in implicit form, and use this to plot the graph of the solution, or to obtain numerical approximations.

If dsolve says 'Explicit solution could not be found' there are two possibilities: (Note that different versions of the symbolic toolbox behave differently)

1. dsolve returns the answer in the form RootOf(expression,z) or solve(equation,y)
   Example 1: Solve the IVP y'=t/(y⁴-1), y(1)=0.
   dsolve('Dy=t/(y^4-1),y(1)=0','t')
   returns in Matlab R2010b
   RootOf(X89^5 - 5*X89 - (5*t^2)/2 + 5/2, X89)
   This means that the solution in implicit form is
   y⁵ - 5y - 5t²/2 + 5/2 = 0
2. dsolve returns the answer [ empty sym ]
   In this case Matlab was unable to find the solution in implicit form. In older versions (e.g. Matlab R2010b) this can even happen when it easy to find by hand the solution inimplicit form. In some cases omitting the initial condition helps:
   For Example 1 newer Matlab versions (R2011b, R2012b) return [empty sym]. In thiscase using dsolve('Dy=t/(y^4-1)','t') gives the implicit solution with a constant.We can then find the value of the constant using the initial condition.

Plotting the solution of IVP in implicit form: If the solution in implicit formis expression=0 use
ezplot(expression,[ t_min t_max y_min y_min ])
to plot the solution y(t) for t_min ≤ t ≤ t _max, y

Matlab R2012b With Usb

_min ≤ y ≤ y _max.
E.g., for Example 1 we can plot the initial point together with the solution curve by
hold on; plot(1,0,'o');
ezplot('y^5 - 5*y + 5/2 - 5*t^2/2',[-2 2 -2 2]); grid on; hold off

We see from the graph that the interval where the solution exists is roughly (-1.6, 1.6).

Plotting the general solution in implicit form: If the general solution in implicit formis expression=C with C arbitrary, use
ezcontour(expression,[ t_min t_max y_min y_min ])
E.g., for Example 1 we can plot the general solution by
ezcontour('y^5 - 5*y + 5/2 - 5*t^2/2',[-2 2 -2 2])
ezcontour plots contours for 9 values of C. If you want to see more contour curves: Downloadthe file ezcontourc.m and put it in the same directory as your otherm-files. Then you can use e.g.
ezcontourc('y^5 - 5*y + 5/2 - 5*t^2/2',[-2 2 -2 2],50)
to obtain contour curves for 50 values of C.

Matlab R2012b With Crack

Finding values of the solution in implicit form:
For Example 1 we obtained the solution in implicit form y⁵ - 5y + 5/2 - 5t²/2 = 0.
We now want to find y(1.5): We plug t=1.5 into the equation and need to solve the equationy⁵ - 5y + 5/2 - 5·1.5²/2 = 0 for y. From the graph above we can see that there are actually 3 solutions: near -1.5, near -0.5, and near 1.5. The solution we want is the one near -0.5.

To find a solution y near -0.5 use
t=1.5; fzero(@(y)y^5-5*y+5/2-5*t^2/2,-0.5)
which returns the answer y=-0.647819.

Tobias von Petersdorff