how to solve this differential equation system symbolically?

Answer 1

[John D'Errico]

This seems like a homework problem. But Walter has already given an answer, that I think is not helpful, since dsolve will fail when applied directly. And since the problem is insufficiently defined to be a good homework problem, I might conjecture it is not in fact homework.

syms y(t) x(t) f(t)

dy = diff(y);
E1 = x + dy == f;
E2 = diff(x) + y == diff(f);

First, differentiate E1, subtracting E2 from that.

E3 = simplify(diff(E1) - E2)

E3(t) = 

As you can see, x(t) and f(t) have been completely eliminated from the result. However, now we can solve for y(t).

y(t) = dsolve(E3,y(0) == 0, dy(0) == 1)

y(t) = 

In fact, if you think about it, note that this is a well known function : the hyperbolic sine function.

y = rewrite(y,'sinh')

y(t) = 

Now that we have y(t), return to equations E1 and E2.

E1 = subs(E1)

E1(t) = 

From this we see that x(t) and f(t) are simply related, differening only by that cosh(t) term.

E2 = subs(E2)

E2(t) = 

The problem is, at this point you are now stuck. Lacking f(t) or y(t), you cannot find the other. What is worse, We see that now E1 and E2 are actually the same pieces of information, since we can derive E2 from E1. That is, just differentiate both sides of E1, and look to see what you get.

diff(E1)

ans(t) = 

SURPRISE! Differentiating E1 yields E2.

The point is, you lack sufficient information to solve the problem, even though you can solve for y(t). So perhaps that is the point of this assignment, IF it is one. Anyway, since the problem already had an answer, I might as well have written this.

[Walter Roberson]

The approach I used worked fine. I tested with MATLAB Online before I posted. I didn't post the code because it is obvious that it is an assignment.

Answer 2

[Walter Roberson]

Create the symbols and the two equations. dsolve() the pair of equations without boundary conditions, getting out a struct with a definition for x and y. Take the y and substitute t = 0 to get y(0) and equate that to the known value. Take the y and differentiate and substitute t = 0 to get y'(0) and equate that to the known value. You now have a pair of simultaneous equations relating the constants of the differential equations.

[Paul]

The two equations have three derivatives. How does dsolve decide which two functions to solve for?

[John D'Errico]

You cannot solve the problem if f(t) is unknown. Or perhaps I should say, if at least one of f(t) or x(t) are not known. In fact, dsolve will tell you the problem is insufficiently determined.

[Walter Roberson]

You cannot solve the problem if f(t) is unknown

Oh?

syms y(t) x(t) f(t)
dy = diff(y);
E1 = x + dy == f;
E2 = diff(x) + y == diff(f);
sol = dsolve(E1, E2)

sol = struct with fields:
    x: exp(t)*(C1 + (exp(-t)*f(t))/2) + exp(-t)*(C2 + (exp(t)*f(t))/2)
    y: exp(-t)*(C2 + (exp(t)*f(t))/2) - exp(t)*(C1 + (exp(-t)*f(t))/2)

you can now proceed as I described to find the values of C1 and C2, leading you to x(t) and y(t) expressions in terms of exp()'s and f(t)

[Walter Roberson]

@Paul

dsolve() looks for derivatives. Any function that only occurs "plain", with no derivative of the function, is treated sort of like a constant

[Paul]

But diff(f) is in the equations and is not "plain", so I'm curious why dsolve() chose to solve for x and y, instead of x and f, or y and f, if dsolve() could have done so.

[Walter Roberson]

Good question. I wonder if it used symvar or equivalent and choose the first two. I seem to recall that solve() chooses variables starting from x y z (or X Y Z) first