solving ode equation using ode45

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nadeeshani Haggalla 2021-2-18

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https://ww2.mathworks.cn/matlabcentral/answers/749142-solving-ode-equation-using-ode45

编辑： David Goodmanson 2021-2-20

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I have an ode which I need to solve with respect to the time variable.

I have 2 seperate scripts, one contains the function and the other contains the command.

my function is,

===================================

function dy_dt=func(t,y)
dy_dt=0.003445*((y-0.072)/y^3)^0.5;
end

===================================

In the other script I have,

===================================

clc, clear, close all
time=0:0.05:8.5;
[t,y]=ode45(@myfun1,time,0.073);
plot(t,y(:,1));

==================================

When the initial value for y=0.072 I get the following graph;

because y=0.072 leads to an erroneous graph I decided to give y=0.073 as the initial value for which I got the following graph;

This is also erroneous as this is not the result I expect. The graph I expect is attached below.

Can someone assist me on what errors I have done in my code? Thank you very much.

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James Tursa 2021-2-20

Because OP wrote:

"... initial value for y=0.072 ..."

Some clarification from OP would help here, as well as some info on where this differential equation comes from.

David Goodmanson 2021-2-20

I agree about clarification from the OP, who wrote "When the initial value for y=0.072 I get the following graph" but who also later wrote "I decided to give y=0.073 as the initial value" so it did not appear to me that the initial value = y0 was set in stone.

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回答（1 个）

David Goodmanson 2021-2-19

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/749142-solving-ode-equation-using-ode45#answer_627629

编辑：David Goodmanson 2021-2-20

Hi nadeeshani,

UPDATED ANSWER

This is a speculative answer, but perhaps you did not take the square root when you calculated the constant. At any rate, plugging in sqrt(.003445) = 0.0587 in place of .003445 (NOTE: and using initial value 0.073) gives a plot that is very close to the expected one.