I am attempting to solve a fairly simply pair of ODEs that represent a flow field for a fluid dynamics problem. I am attempting to find the particle path but I am getting stuck trying to find x and y first. The ODEs are:
$$ \frac{dx}{dt} = U $$ $$ \frac{dy}{dt} = Usin(x-Ut) $$
... where U is any arbitrary constant.
However, I am continually running into this situation where my constant of integration ends up equaling an expression that, when plugged in, results in 0=0. This is a true statement but it does not allow me to approach the actual equations I'm looking for. For example, the result I get from solving our first equation:
$$ \frac{dx}{dt} = U $$ $$ dx = U \cdot dt$$ $$ \int dx = \int U \cdot dt$$ $$ x = Ut+c$$
Where we are negating the constant on the LHS because subtracting it still results in a constant on the RHS. So, solving for our constant c I then get:
$$ x = Ut + c \implies c = x-Ut$$
But when plugging that in we get:
$$ x = Ut + x-Ut$$ $$x-x = Ut-Ut$$ $$0 = 0$$
The initial conditions given are:
$$(x_0, y_0); t = t_0$$
I'm a bit lost here. What part of the equation am I looking for my final x and y, which I can then use to find the path? Plugging in the initial conditions at any point still results in 0 = 0, so I suspect this is where I am going wrong. The same thing happens when I do this with the second equation. I feel as though I am missing an extremely simple thing, but I am a bit fuzzy on the solving.
Since the additive constant $c$ vanishes on derivation, every function of the form $x(t) = Ut + c$ solves your differential equation $\frac{dx}{dt} = U$. So in order to determine $c$, you would need an aditional bit of constraint or information. For example, an initial value $x(t_0)$ will do.
Substituting an equation into itself unsurprisingly gives you $0 = 0$ and that has nothing to do with calculus. From any equation $y = z$, you can subtract $y$ from the left side and $z$ from the right side (as they're equal) and get $0 = 0$. This is equivalent to what you did.