# Laplace transform of a function squared, help with this system

## Homework Statement

Use Laplace transform to the system:

$\frac{dy}{dt} + 6y = \frac{dx}{dt}3x - \frac{dx}{dt} = 2\frac{dy}{dt}$

$x(0) = 2 ; y(0) = 3$

## The Attempt at a Solution

I've tried everything on this one. I first solved $\frac{dy}{dt} + 6y = 2\frac{dy}{dt}$ and I got $y = 3e^{6t}$.

Next I tried writing it:

$36e^{6t} = 3 \frac{d}{dt}(\frac{x^2}{2}) - \frac{dx}{dt}$ so that I could use the identity of the laplace transform of derivatives. That still leaves me with trying to find the transform of x2(t)...

So then I tried

$36e^{6t} dt = 3x - 1 dx$

and integrating, but this brings me to the same problem.

I can't either figure out how to solve it without using laplace transform, so I'm really stuck. What am I doing wrong???

Last edited:

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Use Laplace transform to the system:

$\frac{dy}{dt} + 6y = \frac{dx}{dt}3x - \frac{dx}{dt} = 2\frac{dy}{dt}$

$x(0) = 2 ; y(0) = 3$

## The Attempt at a Solution

I've tried everything on this one.

Everything except what you were asked to do. Start by taking the Laplace transforms of the original equations to get equations involving ##X(s)## and ##Y(s)##.

Ok, I'm still not sure how that changes anything...

$(s+6)Y(s) - 3 = 3L(\frac{dx}{dt}x) - sX(s) +2 = 2sY(s) -6$

I could solve for that middle transform, but replacing it into another equation will just give me 0=0...

What now?

LCKurtz
You have the "system" as$$\frac{dy}{dt} + 6y = \frac{dx}{dt}3x - \frac{dx}{dt} = 2\frac{dy}{dt}$$
I read that to mean this pair of equations:$$\frac{dy}{dt} + 6y = \frac{dx}{dt}$$ $$3x - \frac{dx}{dt} = 2\frac{dy}{dt}$$