Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz5 => Topic started by: Victor Ivrii on November 18, 2018, 04:21:35 AM

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
$$\left\{\begin{aligned}
& x'_1= 0.5x_1 + 2x_2, &&x_1(0) = 2,\\
&x'_2= 2x_1  0.5x_2, &&x_2(0) = 2
\end{aligned}\right.$$

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:
$x_1 = e^{0.5t}(2\cos(2t) + 16\sin(2t))$
$x_2 = e^{0.5t}(0.5\cos(2t) + 10\sin(2t))$

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))

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

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 $x_2=e^{0.5t}(2cos(2𝑡)+2sin(2𝑡))$

This is my complete answer.
It is much easier to simplify the two equations into the common format first.