MATLAB Answers

0

ODE45, differential equation

Asked by Takey Asaad on 14 Sep 2018 at 23:29
Latest activity Edited by Stephan Jung on 17 Sep 2018 at 13:17

my function is

dy/dt=k*y*exp(450/y)

k is constant and y(0)=40 and y(15)=95 solve this equation by using ode45 can someone pleaseeeeeeeeeee check the code and make it work .

tspan = [0 300];
y0 = 40;y15=95
[t,y] = ode45(@(t,y) 'k'*y*exp(450/y), tspan, y0,y15);
plot(t,y,'-o')

  0 Comments

Sign in to comment.

4 Answers

Answer by Stephan Jung on 16 Sep 2018 at 20:39
Edited by Stephan Jung on 17 Sep 2018 at 10:05
 Accepted Answer

Hi,

if you search a value for k that complies with the boundary conditions, you could use fzero to solve the problem numerically:

k = fzero(@calculate_k, 0.001);
disp(['k = ', num2str(k)])
[t,y] = ode45(@(t,y) k*y*exp(450/y), [0 300], 40);
plot(t,y,'r')
hold on
scatter(15,95,'ob')
text(23,96,'y(t=15) = 95','Color','b')
hold off
function k_value = calculate_k(x)
tspan = [0 15];
y0 = 40;
[~,y] = ode45(@(t,y) x*y*exp(450/y), tspan, y0);
k_value = y(end) - 95;
end

This will give you as result:

k = 0.00010435

or if you need it more accurate:

k =
       1.043495007807761e-04

and the plot which belongs to this result looks like the boundary conditions are met (To be precise, it keeps the boundary conditions. This is because fzero is a powerful tool and also because you can think of a very good initial value for x0 with just a few estimates...):

.

Best regards

Stephan

  2 Comments

Thanks so much

No problem - please be patient with your questions. Dont ask the same question a second time, if it needs more time than you expected to get an answer.

Better try to give as much as and as good as possible informations that are needed to let the contributers give you a useful answer.

Sign in to comment.


Answer by James Tursa
on 15 Sep 2018 at 0:29

Remove the quotes from 'k', and be sure to define k before you call ode45. Also, ode45 is an initial value problem solver, so the y15 variable is not applicable (remove it from the call).

  9 Comments

function bvp4
xlow=0;xhigh=300;
solinit=bvpinit(linspace(xlow,xhigh,300),[40 95]);
sol=bvp4c(@bvp4ode,@bvp4bc,solinit);
xinit=linspace(xlow,xhigh,20);
sxint=deval(sol,xint);
plot(xint,sxint(1,:);
function dydx=bvp4ode(x,y)
dydx=[y(2) a*y*exp(450/y);
    function res=bvp4bc(ya,yb)
    res=[ya(1)-1,yb(1)
function bvp4
xlow=0;xhigh=300;
solinit=bvpinit(linspace(xlow,xhigh,300),[40 95]);
sol=bvp4c(@bvp4ode,@bvp4bc,solinit);
xinit=linspace(xlow,xhigh,20);
sxint=deval(sol,xint);
plot(xint,sxint(1,:);
function dydx=bvp4ode(x,y)
dydx=[y(2) a*y*exp(450/y);
    function res=bvp4bc(ya,yb)
    res=[ya(1)-1,yb(1)
 function bvp4
 ↑
Error: Function definition not supported in this context. Create functions in code file.
function bvp4
xlow=0;xhigh=300;
solinit=bvpinit(linspace(xlow,xhigh,300),[40 95]);
sol=bvp4c(@bvp4ode,@bvp4bc,solinit);
xinit=linspace(xlow,xhigh,20);
sxinit=deval(sol,xint);
plot(xinit,sxinit(1,:);
function dydx=bvp4ode(x,y)
dydx=[y(2) ,a*y*exp(450/y)];
    function res=bvp4bc(ya,yb)
    res=[ya(1)-1,yb(1)]

Sign in to comment.


Answer by Takey Asaad on 15 Sep 2018 at 2:04

if true
  function bvp4
xlow=0;xhigh=300;
solinit=bvpinit(linspace(xlow,xhigh,300),[40 95]);
sol=bvp4c(@bvp4ode,@bvp4bc,solinit);
xinit=linspace(xlow,xhigh,20);
sxinit=deval(sol,xint);
plot(xinit,sxinit(1,:);
function dydx=bvp4ode(x,y)
dydx=[y(2) ,a*y*exp(450/y)];
  function res=bvp4bc(ya,yb)
  res=[ya(1)-1,yb(1)]
end

  0 Comments

Sign in to comment.


Answer by Takey Asaad on 15 Sep 2018 at 2:12

code

  0 Comments

Sign in to comment.