System of Differential Equations

[Shelnutt2]

Homework Statement

So my friend asked me for help because he assumed having a math degree meant I knew math

[PLAIN]http://courses.webwork.maa.org:8080/wwtmp/equations/0e/c5a05957810916cfdff379ee0642fc1.png

(dx/dt=−4y and dy/dt = −4x)

Convert this system to a second order differential equation in y by differentiating the second equation with respect to t and substituting for x from the first equation.

Solve the equation you obtained for y as a function of t; hence find x as a function of t. If we also require x(0) = 4 and y(0) = 1, what are x and y?

Homework Equations

y(t)=C1e^(at)cos(Bt) + C2e^(at)sin(Bt)

The Attempt at a Solution

So I've tried a few approaches but I've failed. This is the approach I got the farthest with.


d^2y/d^2t = -4

you can then say r^2 + 4 = 0

b^2-4ac = 0 - 4*4 = -16

r = 0 +/- sqrt(-16) / 2 = +/- 2i
y = Ae^(0t)cos(2t) + Be^(0t)sin(2t)
y=Acos(2t)+Bsin(2t)
y' =2Bcos(2t) + 2Asin(2t)

If I solve for x(t) I end up with the same equation as y(t).
If you set up for x(0)=4 and y(0)=1, you end up with:

1=A
4=A?
I know this is easy but I'm just not seeing it. Any help would be appreciated. Thanks

 

[gabbagabbahey]

This is the approach I got the farthest with.


d^2y/d^2t = -4

No, $\frac{d}{dx}\left(\frac{dy}{dt}\right)=-4$ but [tex]\frac{d^2y}{dt^2}=\frac{d}{dt}\left(-4x\right)=-4\frac{dx}{dt}[/itex]

 

[Shelnutt2]

No, $\frac{d}{dx}\left(\frac{dy}{dt}\right)=-4$ but [tex]\frac{d^2y}{dt^2}=\frac{d}{dt}\left(-4x\right)=-4\frac{dx}{dt}[/itex]

Ah ok, I see.

Now I get $\frac{-1}{4}\frac{d^2y}{dt^2}$ = -4y
$\frac{-1}{4}\frac{d^2y}{dt^2}$ +4y = 0
so,
$\frac{-1}{4}r^2 + 4y = 0$

y(t) = C1 e^(4t) + C2e^(-4t)

Now doing the same for x(t), I end up with the same equation of

x(t) = C1 e^(4t) + C2e^(-4t),

however given my initial values that can't be.

 

[gabbagabbahey]

Ah ok, I see.

Now I get $\frac{-1}{4}\frac{d^2y}{dt^2}$ = -4y
$\frac{-1}{4}\frac{d^2y}{dt^2}$ +4y = 0
so,
$\frac{-1}{4}r^2 + 4y = 0$

y(t) = C1 e^(4t) + C2e^(-4t)

Good..

Now doing the same for x(t), I end up with the same equation of

x(t) = C1 e^(4t) + C2e^(-4t),

however given my initial values that can't be.

$x(t)$ is related to $y(t)$ via $y'(t)=-4x(t)$, once you've found $y(t)$, you need only differentiate it once and divide by -4 to obtain $x(t)$

 

[Shelnutt2]

Good..



$x(t)$ is related to $y(t)$ via $y'(t)=-4x(t)$, once you've found $y(t)$, you need only differentiate it once and divide by -4 to obtain $x(t)$

Thank you for leading me down the right path! I got it now :)

The solution was
y(t) = (5/2)e^(-4t)+(-3/2)e^(4t)
x(t) =(5/2)e^(-4t)-(-3/2)e^(4t)