Toronto Math Forum
MAT244--2018F => MAT244--Tests => Quiz-5 => Topic started by: Victor Ivrii on November 18, 2018, 04:21:35 AM
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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
$$\left\{\begin{aligned}
& x'_1= -0.5x_1 + 2x_2, &&x_1(0) = -2,\\
&x'_2= -2x_1 - 0.5x_2, &&x_2(0) = 2
\end{aligned}\right.$$
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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:
$x_1 = e^{-0.5t}(-2\cos(2t) + 16\sin(2t))$
$x_2 = e^{-0.5t}(-0.5\cos(2t) + 10\sin(2t))$
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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:
𝑥1=𝑒−0.5𝑡(−2cos(2𝑡)+2sin(2𝑡))
𝑥2=𝑒−0.5𝑡(2cos(2𝑡)+2sin(2𝑡))
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I think Michael's x2 is wrong. I also think c1 =-2 and c2 =2 same answer with Zihan.
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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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This is my complete answer.
It is much easier to simplify the two equations into the common format first.