Symbolic Solving of 1st and 2nd order ODES

In this section we use maxima to obtain symbolic solutions to certain differential equations. Then at the end we plot these, and also compare with the dfplot trajectories.

We use the ode2 routine: -- Function ode2( differential_equation, dependent_var, independent_var )

To enter a differential, you must put a quote "'" before the diff. This quote stops the derivative from actually being evaluated.

diff(sin(x),x) returns COS(X) and diff(y,x) evaluates to 0

but 'diff(y,x) yields what we want:

                                      dY
                                      --
                                      dX

                                    %C - COS(X)
                                Y = -----------
                                         3
                                        X

                                  1
                                 ---- = X + %C
                                    2
                                 2 Y

                                       2
                            ATAN(Y)   X  + 2 X
                            ------- = -------- + %C
                               2         2

Where is the above solution valid?

Notation: In maxima variables may have multiple letters, so that 'ab' is one variable, and you must write 'a*b' to mean the product of two symbols named 'a' and 'b'. Similarly 'dy' is just a two letter symbol. The printing of dy/dx is a pretty printing for 'diff(y,x) , but you must use this latter representation to enter the form. To get a display which uses the same notation for input and output do
display2d:false . If you do that the last form evals to the more compact but perhaps less human readable ATAN(Y)/2 = (X^2+2*X)/2+%C .

A second order differential equation:

ode2('diff(y,x,2) = 2*(1+x) ,y,x) returns Y = (X^3+3*X^2)/3+%K2*X+%K1

Initial Values:

It is possible to have the solver substitute in initial values, to determine the constant(s), in the above solutions. The functions for doing that are

IC1(sol,x=x0,y=y0)

IC2 is used for specifying an initial tangent and a point to pass through: For example the following forces the slope to be 3 at the point (1,2).

(soln:ode2('diff(y,x,2) = 2*(1+x) ,y,x), tmp:IC2(soln,x=1,y=2,'diff(y,x)=3)) returns

                                    3      2
                                   X  + 3 X    2
                               Y = --------- + -
                                       3       3

After having clicked on the above you may plot this: we set tmp to be the solution equation,

plot2d(sublis(solve(tmp,y),y),[x,-4,4],[y,-10,10])

sol:ode2('diff(y,x) = 2*(1+x)*(1+y**2) ,y,x) returns

                                       2
                            ATAN(Y)   X  + 2 X
                            ------- = -------- + %C
                               2         2

Lets plot this solution for x in [2,4] and y in [-3,6], and verify that the point (3,4) does lie on the particular solution computed by IC1. After you have clicked on the above, to set sol correctly we can plot the particular solution (in order to keep plot2d happy, we have to solve for y) :

plot2d( sublis(solve( IC1(sol,x=3,y=4) ,y),y), [x,2,4],[y,-3,6])

We could look at the following, and click at the position (3,4) to find the solution numerically.

plotdf -ode {d(y,x)=2*(1+x)*(1+y**2)} -yradius 10 -xcenter 3 -xradius 1 evaluates to RESULT

This uses the default 4th order Runge Kutte with a fixed step in the independent variable x. Note that this gives an answer which is sloped a bit too much to the right at the top, whereas our theoretical plot was practically vertical when going through (3,4). We can retry using an 'adaptive 5th order Runge Kutta method (RKQC)'. This alters the x step according to an error estimate, and gives a more accurate picture. NOTE THE (RKQC) IS not yet implemented in netmath.

plotdf -ode {d(y,x)=2*(1+x)*(1+y**2)} -yradius 10 -xcenter 3 -xradius 1 returns RESULT