Question

a)Transform the given system into a single equation of second order.
b)Find x1 and x2 that also satisfy the given initial conditions.
I am having problems with part b), wondering if my a) is right.
Here are the equations:
x1′ =2x2, x1(0)=3
x2′ =−2x1, x2(0)=4

I got x2=c1e^t cos2t + ic2e^tsin2t
and x1=the same thing???
I can get c1 for both as sin2t becomes 0 and e^t becomes 1, so for both x(0)=c1. How do I find c2? Are my solutions wrong?

Answer 1

Equations are \[x_1^{'}=2x_2~and~x_1(0)=3\]\[x_2^{'}=-2x_1~and~x_2(0)=4\]

Answer 2

Yes, those are the equations. It wants me to transform them into one second order de and then solve.

Answer 3

let me differentiate x1' \[x_1^{''}=2x_2^{'}\]so we can now substitute x2' \[x_1^{''}=2(-2x_1)=-4x_1\]\[\frac{ d^2x_1 }{ dt^2 }=-4x_1~~~~~(a)\]

Answer 4

I prefer the use of Laplace Transformation. So for x1, it will be \[x_1^{'}=2x_2\]\[sX_1(s)-x_1(0)=2X_2(s)~~~~substitute~initial~condition\]\[sX_1(s)-3=2X_2(s)\]\[sX_1(s)-2X_2(s)=3~~~~say~eqn.(1)\]and for x2, it will be \[x_2^{'}=-2x_1\]\[sX_2(s)-x_2(0)=-2x_1~~~~substitute~initial~condition\]\[sX_2(s)-4=-2X_1(s)~~~~say~eqn.(2)\]

Answer 5

Thanks, we did not cover Laplace transforms unfortunately.
Based on your differentiation, x1"+4x1=0, which has the roots (+/-)2i, which with eulers formula gives x1= c1e^t cos2t + c2e^t sin 2t, right?
How can I find both constants given the initial conditions?

Answer 6

If the root from the characteristic equation is of the form \(\large a\pm bi\) with \(\large a\neq 0\), then the general solution is
\[\large x_1= e^{at}(c_1\cos(bt)+c_2\sin(bt))\]
However, the root you found is \(\large \pm 2i\); thus, the general solution is of the form
\[\large x_1 = c_1\cos(2t)+c_2\sin(2t).\]
Since \(\large x_1^{\prime}=2x_2\), substitute the \(\large x_1\) you found into this equation. Then apply the initial conditions to find the constants and you'll have the particular solution.

Does this make sense? :-)

Answer 7

ahh ok sorry it's just another method of solving DEs... :)