# How to solve 2nd ODE equation in numerical and analytical method at same plot graph?

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Shreyas Sangamesh 2022 年 9 月 28 日

Hello,
I am was looking for help to solve this equation
a. dx/dt = -x
b. dx/dt = -x+1
in Matlab, can I get help to solve this equation of plot the numerical solutions and analytical solutions in the same graph and compare them.
with ode45
can anyone help me with this code?
##### 6 件のコメント4 件の古いコメントを表示4 件の古いコメントを非表示
Shreyas Sangamesh 2022 年 9 月 28 日
Its not the same one I asked before,
I just wanted to have some assistance on how these equation behave in 1st order ODE,
I have got the revelant idea on how to carry it out!!
Thank you
Torsten 2022 年 9 月 28 日

a .
dx/dt = - x -> dx/x = -dt -> log(x/x0) = (t0-t) -> x = x0 * exp(t0-t)
b.
dx/dt = -x+1 -> dx/(x-1) = -dt -> log((x-1)/(x0-1)) = t0-t -> x-1 = (x0-1)*exp(t0-t) -> x = 1 + (x0-1)*exp(t0-t)

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### 回答 (1 件)

Sai 2022 年 10 月 12 日
I understand that you are trying to solve the differential equations both numerically and analytically and get the plots on same graph for comparison.
You can use available MATLAB functions “ode45” for numerical approach and “dsolve” for analytical approach.
Please refer to the attached code snippets, which can help you solve the problem.
NOTE: Since the initial conditions are not given, they are assumed
a). dx/dt = -x
%Numerical Method
[t,x] = ode45(@(t,x) -x,[0 20],1)
plot(t,x)
hold on
%Analytical Method
syms x(t)
dx = diff(x)
eqn = dx==-x
x(t) = dsolve(eqn,x(0)==1)
t = 0:20
plot(t,x(t))
hold off
legend("Numerical","Analytical")
b). dx/dt = -x+1
%Numerical Method
[t,x] = ode45(@(t,x) -x+1,[0 20],0)
plot(t,x)
hold on
%Analytical Method
syms x(t)
dx = diff(x)
eqn = dx==-x+1
x(t) = dsolve(eqn,x(0)==0)
t = 0:20
plot(t,x(t))
hold off
legend("Numerical","Analytical")
You can also refer to the below links regarding “ode45” and “dsolve” for future references.

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