MAT244--2018F > Quiz-5

Q5 TUT 0501

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Victor Ivrii:
It looks like I missed it

Transform the given system into a single equation of second order and find the solution $(x_1(t),x_2(t))$, satisfying initial conditions
& x'_1= -0.5x_1 + 2x_2, &&x_1(0) = -2,\\
&x'_2= -2x_1 - 0.5x_2, &&x_2(0) = 2

Michael Poon:
Isolate the first equation for $x_2$:

$x_2 = 0.5x_1' + 0.25x_1$

Differentiating both sides we get:

$x_2' = 0.5x_1'' + 0.25x_1'$

Subbing in the above into the second equation:

$x_1'' + x_1 + 4.25x_1 = 0$

Solving for the characteristing eqn gives us:

$r = -0.5 \pm 2i$

So, we now know :

$x_1 = e^{-0.5t}(c_1cos(2t) + c_2sin(2t))$

Subbing the above and its derivative into eqn 2:

$x_2 = e^{-0.5t}(c_1(0.125)\cos(2t) + c_2(1.125)\sin(2t))$

By initial conditions:

$-2 = c_1$

$2 = 0.125c_2$

Final solution:

$x_1 = e^{-0.5t}(-2\cos(2t) + 16\sin(2t))$

$x_2 = e^{-0.5t}(-0.5\cos(2t) + 10\sin(2t))$

Xu Zihan:
For Michael Poon's answer,
after the substitution,

𝑥2 should equal to 𝑒−0.5𝑡(-𝑐1cos(2𝑡)+𝑐2sin(2𝑡))

And C1=-2   C2=2

So the final solution will be:



Guanyao Liang:
I think Michael's x2 is wrong. I also think c1 =-2 and c2 =2 same answer with Zihan.

Yunqi(Yuki) Huang:
Isolate the first equation for$ x_2:x_2=0.5x′_1+0.25x_1 (1)$
Differentiating both sides we get$x′_2=0.5x″_1+0.25x′_1(2)$
Substitute (1) and (2) in the above into the second equation$x″_1+x_1+4.25x_1=0(3)$
Solving (3):r=−0.5±2i
So, we now know $x_1=e^{-0.5t}(c_1cos(2t)+c_2sin(2t))$ and $x2=e^{−0.5t}(-c_1sin(2𝑡)+c_2cos(2𝑡))$
By initial conditions:$c_1 = -2$ and $c_2=2$
Final solution:$x_1=e^{-0.5t}(−2cos(2𝑡)+2sin(2𝑡))$ and $x_2=e^{-0.5t}(2cos(2𝑡)+2sin(2𝑡))$


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