I Solving linear differential equations

1. Aug 24, 2016

tomwilliam2

I have the following linear differential equations:
$A\dot{x} + By = 0$
$C\dot{y} + Dx = 0$
Where x and y are functions of t, and A through D are constants.
I can solve this fairly easily by differentiating the first equation, rearranging, and removing one of the variables, which gives me a fairly straightforward to solve 2nd order differential equation.
However, my textbook says "these are linear differential equations, whose characteristic equation, in terms of the Laplace Operator, is..." and goes on to produce the same characteristic equation as I get. I understand that when the linear equations get more complicated, you have to use a Laplace transform to solve them. So I thought I'd give it a go, but can't get the same characteristic equation. I'm not sure where I'm going wrong, so if I put my working out here, could someone point me to the mistake(s)?
I'm starting with the first equation:
$F(s) = A \int_{0}^{\infty} \dot{x}e^{-st} dt + B \int_{0}^{\infty} ye^{-st}dt$
Using integration by parts:
$F(s) = A\left(xe^{-st} + s \int_{0}^{\infty}xe^{-st} dt \right) + B/s \left( \int_{0}^{\infty}e^{-st} dt - ye^{-st}\right)$
Now applying integration by parts again:
$F(s) = A\left((xe^{-st} -xe^{-st} - 1/s\right) -By/s - B/s^2 \left(\int_{0}^{\infty}e^{-st} dt\right)$
And I think here I have to apply the limits to clear it up, so:
$F(s) = -A/s - By/s + B/s^2 = \frac{B(1-ys)}{s^2} - \frac{A}{s}$
But I think this has already gone wrong somewhere, as I don't see this becoming a quadratic equation.
Presuming I can fix this integration by parts, what is the next step? To do the same thing with the other linear equation and then equate them?
Thanks in advance for any suggestions.

2. Aug 24, 2016

Lucas SV

I don't see where some terms of the second line come from. $xe^{-st}$ needs the be evaluated at the boundary points $0$ and $\infty$, since that's how partial integration works. In the $B$ term, why did the $y$ disappear? The only think I can think of that I'm missing is that you may have used the differential equations.

3. Aug 24, 2016

tomwilliam2

Thanks for your comment. I missed out some of the steps, as it was painstaking to write out, but I might have made a mistake somewhere. If I take just the B term:
$F_B(s) = B \int_{0}^{\infty} ye^{-st} dt$
Using integration by parts, we set
$v = y$
$v' = 1$
$u = -\frac{1}{s}e^{-st}$
$u' = e^{-st}$
and $\int u'v = uv - \int uv'$
So
$B \int_{0}^{\infty} ye^{-st} dt = B \left(-\frac{y}{s}e^{-st} + \frac{1}{s}\int_{0}^{\infty}e^{-st}\right) dt$ evaluated between t=0 and t=infinity.
So:
$F_B(s) = B\left(-\frac{y}{s}e^{-st} + \frac{1}{s^2}e^{-st}\right)$ from t=0 to t=infinity.
$F_B(s) = \frac{By}{s} - \frac{B}{s^2}$
So that's how I got the result for the B term. Did I make a mistake?

4. Aug 24, 2016

Lucas SV

Yes you did, your integration variable is $t$ and $y$ is a function of $t$ as stated in the beggining. So $v'=y'$ is only equal to $1$ if $y$ is a linear function with respect to $t$.

5. Aug 25, 2016

tomwilliam2

Ah, of course. How silly of me. So I'll go back and try to solve it again. I presume that when I get an expression for the Laplace transform of the first equation, I have to then equate it with the transform of the second equation?
That makes me wonder, if f(x) = g(x), then does L(f) = L(g)? I.e. if two linear equations are equal to each other, then are their laplace transforms also equal to each other?
Thanks again.

Last edited: Aug 25, 2016
6. Aug 25, 2016

Lucas SV

Yes this is true. If any two objects in mathematics are equal, then you can substitute one for the other in an arbitrary expression. To be more precise you have to say $f(x)=g(x)$ for all $x\in D$, where $D$ is the common domain of $f$ and $g$ (they need to have common domain in order to be equal).

7. Aug 25, 2016

tomwilliam2

Thanks. So I'm left trying to integrate:
$\int_{0}^{\infty}\dot{x}e^{-st}dt$
Which has got me completely stuck.
I think I'll just leave the LaPlace transform. It's a shame though, because I understand it is a good tool to use in some cases where the differential equations are more difficult than here.

8. Aug 25, 2016

Lucas SV

It's fine, come back at a later stage. Sometimes there are things we do not see straight away when learning, but some time later when we come back to it, it just clicks and it all makes sense.