Finding a Solution to a System of Differential Equations

[transmini]

Homework Statement

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
$$\bf{\phi} (0)=(0,0)$$

Homework Equations

A hint was given to define $v(t)=y_1(t)+y_2(t)$

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.

 

[Mark44]

Homework Statement

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
$$\bf{\phi} (0)=(0,0)$$

Homework Equations

A hint was given to define $v(t)=y_1(t)+y_2(t)$

The Attempt at a Solution

Using the suggested substitution, the system became
$$v'(t)=v(t)+y'_2(t)+f(t)$$
$$v'(t)=v(t)+y'_1(t)$$

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.

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.

 

[transmini]

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.

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.

 

[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?

 

[Mark44]

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.

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

 

[transmini]

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

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.

 

[ehild]

I then added these together to get
$$v'(t)-2v(t)=f(t)$$ where $$v'(t) = y'_1(t)+y'_2(t)$$

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.

 

[transmini]

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.

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

 

[ehild]

Great, good job !