# System of ODE's

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1. Sep 10, 2016

### transmini

1. The problem statement, all variables and given/known data
Find a solution $\bf{\phi}$ of the system
$$y'_1(t)=y_1(t)+y_2(t)+f(t)$$
$$y'_2(t)=y_1(t)+y_2(t)$$
where f(t) is a continuous function
$$\bf{\phi} (0)=(0,0)$$

2. Relevant equations
A hint was given to define $v(t)=y_1(t)+y_2(t)$

3. The attempt at a solution
Using the suggested substitution, the system became
$$v'(t)=v(t)+y'_2(t)+f(t)$$
$$v'(t)=v(t)+y'_1(t)$$

I then added these together to get
$$v'(t)-2v(t)=f(t)$$ where $$v'(t) = y'_1(t)+y'_2(t)$$
This was solved using an integrating factor to get
$$v(t)=e^{2t} \int_0^t e^{-2t}f(t)dt$$

but then from here I have no idea where to go, if this was even the proper path to begin with.

Last edited by a moderator: Sep 10, 2016
2. Sep 10, 2016

### Staff: Mentor

It's not immediately obvious to me how you got the system above.
If $v(t) = y_1(t) + y_2(t)$, then $v'(t) = y_1'(t) + y_2'(t)$. I don't see how you got either of the equations just above.

3. Sep 10, 2016

### transmini

The first equation was, as listed under the relevant equations section, given as a hint out of the book. The latter was found by taking the derivative with respect to t of both sides of the first.

4. Sep 10, 2016

### Twigg

You can't solve the system as is because you only have one known and two unknowns. You need to find another substitution that is linearly independent of v(t) that results in an integrable ODE. Hint: what looks a lot like $v(t) = y_{1}(t) + y_{2}(t)$ but is much easier to solve for?

5. Sep 10, 2016

### Staff: Mentor

I get that, but how did you get this system?
$$v'(t)=v(t)+y'_2(t)+f(t)$$
$$v'(t)=v(t)+y'_1(t)$$

6. Sep 10, 2016

### transmini

Oh I misread that post, since
$$v'(t) = y'_1(t)+y'_2(t)$$
I solved that equation for $y'_1(t)$ and $y'_2(t)$ substituted them into the original systems. That step didn't seem very helpful when doing though since all it does is switch which derivative is in which equation.

7. Sep 11, 2016

### ehild

If you add the two initial equations, you get v'(t)=2v(t)+f(t) which has the solution you have derived: $v(t)=e^{2t} \int_0^t e^{-2τ}f(τ)dτ$
(The integrating variable should not be the same as the limit of integration.)
You can also introduce the other new unknown function (as suggested by @Twigg )w(t)=y1(t)-y2(t), and get the differential equation for w(t): w'(t) = f(t).
If you know v(t) and w(t), you can express y1(t) and y2(t) in terms of them.

Last edited: Sep 11, 2016
8. Sep 11, 2016

### transmini

This worked great. After combining the systems both through addition for one set of equation and subtraction for another set, it led me to the the $v(t)$ I had and $$w(t) = \int_0^t f(\tau)d\tau$$

Using linear algebra to solve the system comprised of $v(t)$ and $w(t)$ gave me
$$y_1(t) = \frac{1}{2}(v(t)+w(t)$$
$$y_2(t)=\frac{1}{2}(v(t)-w(t))$$

which gave me the solution that satisfies both the system and the initial conditions of
$$y_1(t)=\frac{1}{2}e^{2t}\int_0^te^{-2\tau}f(\tau)d\tau+\frac{1}{2}\int_0^tf(\tau)d\tau$$
$$y_2(t)=\frac{1}{2}e^{2t}\int_0^te^{-2\tau}f(\tau)d\tau-\frac{1}{2}\int_0^tf(\tau)d\tau$$

Thanks to both @Twigg and @ehild for leading me towards this answer

9. Sep 11, 2016

### ehild

Great, good job !