# I Problems with this integral

1. Jun 22, 2017

### MikeSv

Hello everyone.

Iam trying to get my head around a solution for an integral but I can't figure out how its done.

I have given the following :

x1'(t) = 0
x2'(t) =tx1(t)

Where " ' " indicates the derivative.

Talking the time integral the result is given by:

x1(t) = x1(t0)
x2(t) = x2(t0) + 1/2(t^2-t0^2)x1(t0)

It would be great if anyone could help me out or give me a hint.

Cheers,

Mike

2. Jun 22, 2017

### Staff: Mentor

Your solution looks fine to me, although it's more complicated than it needs to be with all the subscripts.
For simplicity in writing, I'm going to rephrase your problem:

x' = 0
y' = tx

Here, both x and y are functions of t.

Since x' = 0, then $x = k_1$, for some constant $k_1$. After substitution into the second equation, you get
$y = \frac 1 2 k_1t^2 + k_2$
It's always a good idea to verify that your solution actually works, by substituting back into the original system of equations.

Since there are no initial conditions given (or at least shown here), we're done.

Note that this is a very simple system of differential equations, one that can be "uncoupled" by substitution. More complicated systems, in which each derivative is in terms of the other function, require much more complicated techniques.

3. Jun 22, 2017

### MikeSv

Hi and thanks for the reply

I guess I just got confused by the subsribts.
And if I see correctly the integral is evaluated from t0 to t.

Cheers,

Mike

4. Jun 22, 2017

### Staff: Mentor

No need to use definite integrals.
$x'(t) = 0 \Rightarrow x(t) = \int 0 dt = k_1$
$y'(t) = t k_1 \Rightarrow y(t) = k_1 \int t~ dt = \frac 1 2 k_1 t^2 + k_2$

Of course, since the problem is in terms of x1 and x2, the solutions should be as well.

5. Jun 22, 2017

### MikeSv

Thank you so much for clarifying!
This helped a lot,!

Cheers,

Mike