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MATLAB second order differential equation & system examples
Engineering 240
March 12, 2013
1. Solve the boundary value problem:
d2 y
dx2
−
dy
dx
− 2y = 0 and y(0) = 1 and y 0 (3) = 2
First transform the equation to a system:
(
dy1
dx
dy2
dx
=
y2
= 2y1 + y2
)
with y1 (0) = 1 and y2 (3) = 2
Create a function file for the derivative:
function ydot=yprime(t,y)
ydot=[ y(2) ; 2*y(1) + y(2)];
end
Create a function file for the boundary value condition errors:
function res = bc(ya,yb)
res=[ ya(1)−1 ; yb(2)−2 ]
end
Initialize the x-values with step size 1/10 and provide a guess for the initial y and y 0 values:
solg = bvpinit( linspace(0,3,31) , [ 1 1 ] );
% this is a built-in function which can be called from the
command line
Generate a solution to the boundary value problem:
sol = bvp4c(@yprime, @bc, solg);
Graph the approximate solution y and the approximate solution’s derivative y 0 at the evaluation points with step-size h = 1/10:
xx=sol.x ; yy = sol.y ; yder=sol.yp ;
plot( xx, yy , ‘k’ , xx, yder , ‘g’ )
Experiment with the same differential equation but with different boundary conditions:
• y(0) = 5 and y(3) = −1
• y 0 (0) = 0 and y 0 (3) = 1
2. Find the approximate solution to Lorenz’ equations:
(first used to model atmospheric convection!)
 dx

 dt
dy
dt

 dz
dt
=
σ(y − x)
= x(ρ − z) − y
=
xy − βz





Create a function file:
function ydot=lz(t,y);
sigma=10; rho=28; beta=8/3;
ydot=[ sigma*( y(2)−y(1) ) ; y(1)*(rho−y(3)) − y(2) ; y(1)*y(2)−beta*y(3)];
end
Generate an approximate solution:
[t y] = ode45(‘lz ’ , [0 40], [ 1 ; 0 ; 0 ] ) ;
Plot the parametric solution:
figure(3)
plot3(y(:,1),y(:,2),y(:,3),’k’)
Experiment with the shape of the solution curve by varying the initial conditions and by
varying the values of the parameters sigma, rho, and beta.
3. A few numerical derivatives
For the function y = f (x) = sin(x), with derivative f 0 (x) = cos(x), compare the exact first
derivative yp with the approximate “centered difference quotient” ypa:
x = 0 : .01 : 3 ;
h = .01 ;
y = sin (x) ; yp = cos(x) ;
ypa = 1/(2*h)*(diff(x(1:end-1)) + diff(x(2:end)) ) ;
plot ( x(2:end-1) , ypa , ‘g’, x(2:end-1) , yp(2:end-1) , ‘k’)
second derivatives:
ypp = -sin(x); % exact
yppa = diff(x,2)/h∧2 ; % approximate
plot ( x(2:end-1) , yppa , ‘g’, x(2:end-1) , ypp(2:end-1) , ‘k’)