Solve second order differential equation with independent variable

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Abhivyakti
Abhivyakti 2014-1-28
评论： Azzi Abdelmalek 2014-1-28
I need the solution to the following second order differential equation:
D2y+(2/x)*Dy= 10*(y/(1+y));
I want to use ode45 for this. I have split the equation in the following manner :
%y(1)=y(x);
%y(2)=Y'(x);
So the equation can be written as :
y'(1)=y(2);
y'(2)=10*(Y(1)/(1+(y(1)))-(2/x)*Y(2);
When I solve this usind ode45, I get a matrix comprising of Nan values.
I believe the variation of the independent variable, i.e x, cannot be done like this.
I am not sure how to solve this.
Any help would be appreciated. Thanks! :)
  2 个评论
Amit
Amit 2014-1-28
Post the code for your function? Also what are the initial values for Y and Y'
Abhivyakti
Abhivyakti 2014-1-28
I just know one condition here, i.e When x is 'R', y id 's0'. Here R,s0 are constants.

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回答（1 个）

Azzi Abdelmalek
Azzi Abdelmalek 2014-1-28
编辑：Azzi Abdelmalek 2014-1-28
Your function should be:
function dz=myode45(x,z)
dz(1,1)=z(2)
dz(2,1)=-2*z(2)/x+10*z(1)/(z(1)+1)
%Call myode45
tspan=[0.1 10];
% don't start tspan at 0, because in your equation there is -2z(2)/x
y0=[0;1]; % initial conditions
[x,y]=ode45('myode45',tspan,y0)
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Abhivyakti
Abhivyakti 2014-1-28
Alright. Tell me one thing, is it that 'tspan' is supposed to incorporate the variation of the values of 'x' ? I understood the values 'y' is storing but what about 'x' and 'tspan' ?
Azzi Abdelmalek
Azzi Abdelmalek 2014-1-28
tspan=[t0 tf] % t0 is a start time, and tf is a final time
x represent time or whatever you want. It's in your equation

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