How do you use ode45( ) when the equation is not in dy/dt form?

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I have an equation that I need to approximate the solution for using ode45( ) with initial values y(0) = 1 and y'(0) = 0 on interval of 0 < t < 150 and then plot x(t)
So far I have this for the equation y" + 4y = sin(1.9t)
syms y(t)
eqn = diff(y,2) + 4*y == sin(1.9*t)
V = odeToVectorField(eqn)
I know if the equation was in the form of dy/dt = sin(1.9t) - 4y then it would be
k = @(t,y) sin(1.9*t)-4*y;
[t,y] = oder45(k,[0,150])
plot(t,y)
But I do not know know how to write it for the form that my equation is in at this moment

Answers (2)

Paul
Paul on 19 Nov 2021
Check out this link ....
  2 Comments
Paul
Paul on 19 Nov 2021
It's shown on the the doc page at that link. Or check
doc ode45
to see how to specifcy the initial conditions

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Star Strider
Star Strider on 19 Nov 2021
Since the equation as written involves the second derivative, it is also necessary to define the first derivative as a separate element of the resulting system of first-order differential equatiions.
The odeToVectorField function does this (I added the second ‘substitutions’ output to illustrate what the function actually does) —
syms y(t)
eqn = diff(y,2) + 4*y == sin(1.9*t)
eqn(t) = 
[V,S] = odeToVectorField(eqn)
V = 
S = 
This is actually
The function presented to ode45 would then be coded using the matlabFunction function.
Explore that process!
.
  10 Comments
Star Strider
Star Strider on 23 Nov 2021
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.

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