There are usually more than one options how to get solve a problem. It is particularly true for math. Let's look on a spring and its mathematical model.

There are many sites describing how to get model of a spring. We will use one of the simplest:

\begin{aligned} \ddot{x} = - \frac{k}{m} * x \qquad (1) \end{aligned}

Let:

k = 2 (spring constant)

m = 1 (we can say that we assume unit mass)

So our differential equation is:

\begin{aligned} \ddot{x} = -2x \qquad (2) \end{aligned}

Let's solve the equation (2), that is let's find x(t) (x function of time) that satisfies the eq. (2).

**Option 1**

We may guess that x(t) must be something special that does not change much when derivated. $e^t$ is the case:

\begin{aligned} \frac{d}{dt}e^t = e^t \qquad (3) \end{aligned}

Then we can write:

\begin{aligned} x(t) = e^{ht} * w \qquad (4) \end{aligned}

where w - initial condition/constant

\begin{aligned} \dot{x(t)} = h * e^{ht} * w \qquad (5) \end{aligned}

\begin{aligned} \ddot{x(t)} = h^2 * e^{ht} * w \qquad (6) \end{aligned}

Now we can plug (4) and (6) into (2):

\begin{aligned} h^2 * e^{ht} * w = -2 * e^{ht} * w \qquad (7) \end{aligned}

Since $ e^{ht} * w \neq 0 $ we can divide both sides by that so we get:

\begin{aligned} h^2 = -2 \qquad (8) \end{aligned}

\begin{aligned} h = \pm \sqrt{-2} = \pm \sqrt{2} * j \qquad (9) \end{aligned}

Note:

h without j => overdumped system

h with j => underdamped system (with oscillations)

We got 2 following solutions:

\begin{aligned} x_{1_{(t)}} = e^{\sqrt{2}jt} * w \qquad (10) \end{aligned}

\begin{aligned} x_{2_{(t)}} = e^{-\sqrt{2}jt} * w \qquad (11) \end{aligned}

Now it is time for famous **Euler's formula**:

\begin{aligned} e^{i * something} = cos(something) + j * sin(something) \qquad (12) \end{aligned}

\begin{aligned} e^{-i * something} = cos(something) - j * sin(something) \qquad (13) \end{aligned}

Our 2 solutions can be written also as:

\begin{aligned} x_{1/2_{(t)}} = cos(\sqrt{2}t) \pm j * sin(\sqrt{2}t) * w \qquad (14) \end{aligned}

Let's use principle of **Superposition - **if I have 2 solutions, I can add/subtract them and scale them. So if we take 2 solutions from (14) and multiply both by 1/2 and add them we just get $cos(\sqrt{2}t) * w$. If we multiply both by -1/2j and subtract them, we get $sin(\sqrt{2}) * w$. These 2 terms combined give us general solution:

\begin{aligned} x_{t} = C_1 * cos(\sqrt{2}t) + C_2 * sin(\sqrt{2}t) \qquad (15) \end{aligned}

Another form of general solution would be:

\begin{aligned} x_{t} = C_1 * e^{\sqrt{2}jt} + C_2 * e^{-\sqrt{2}jt} \qquad (16) \end{aligned}

**IVP (Initial Value Problem) v1:**

We are given following values:

a/ $x_0 = 3$

b/ $\dot{x_0} = v_0 = 4$

And our goal is estimate C1, C2 constants, so we choose 1 particular solution from the general solution.

1. Let's use the first of peace of information which is, the initial position $x_0 = 3$. We can directly plug this into (15):

\begin{aligned} x_{t=0} = C_1 * cos(\sqrt{2}* 0) + C_2 * sin(\sqrt{2}* 0) = 3 \qquad \end{aligned}

\begin{aligned} C_1 * 1 + C_2 * 0 = 3 \qquad \end{aligned}

\begin{aligned} C_1 = 3 \qquad \end{aligned}

To get C2 we have to take derivative of the (15):

\begin{aligned} \frac{dx}{dt} = 4 = - C_1 * \sqrt{2} * sin(\sqrt{2}* 0) + C_2 * \sqrt{2} * cos(\sqrt{2}* 0) \qquad \end{aligned}

\begin{aligned} 4 = C_2 * \sqrt{2} * cos(0) \qquad \end{aligned}

\begin{aligned} 4/ \sqrt{2} = C_2 \qquad \end{aligned}

**IVP (Initial Value Problem) v2:**

We are given following values:

a/ $x_0 = 3$

b/ $\dot{x_0} = v_0 = 4$

Now we can plug $x_0 = 3$ into (16):

\begin{aligned} x_{t} = C_1 * e^{\sqrt{2} * j * 0} + C_2 * e^{-\sqrt{2} * j * 0} = 3 \qquad \end{aligned}

\begin{aligned} C_1 + C_2 = 3 \qquad \end{aligned}

\begin{aligned} C_1 = 3 - C_2 \qquad \end{aligned}

Now let's take derivative of (16):

\begin{aligned} \frac{dx}{dt} = 4 = \sqrt{2}j * C_1 * e^{\sqrt{2} * j * t} + \sqrt{2}j * C_2 * e^{-\sqrt{2} * j * t} \qquad \end{aligned}

\begin{aligned} \frac{dx}{dt} = 4 = \sqrt{2}j * C_1 - \sqrt{2}j * C_2 \qquad \end{aligned}

\begin{aligned} \frac{dx}{dt} = 4 = \sqrt{2}j * (3 - C_2) - \sqrt{2}j * C_2 \qquad \end{aligned}

\begin{aligned} \frac{dx}{dt} = 4 = \sqrt{2}j * 3 - C_2 * \sqrt{2}j - C_2 * \sqrt{2}j \qquad \end{aligned}

\begin{aligned} 4 = 3 * \sqrt{2}j - 2 * C_2 * \sqrt{2}j \qquad \end{aligned}

\begin{aligned} C_2 = \frac{ 3 * \sqrt{2}j - 4}{2 * \sqrt{2}j} \qquad \end{aligned}

\begin{aligned} C_1 = 3 - \frac{ 3 * \sqrt{2}j - 4}{2 * \sqrt{2}j} \qquad \end{aligned}

**IVP (Initial Value Problem) v3 - Laplace transformation:**

We are again given following values:

a/ $x_0 = 3$

b/ $\dot{x_0} = v_0 = 4$

Now we can try to use the Laplace transformation and verify our previous results. I found good explanation of the Laplace transformation here.

The first step is to take Laplace transform of the both sides.

We have to realize that x is function of t -> x(t). And $L\{x(t)\} = X(s) = X$.

\begin{aligned} L\{\ddot{x}\} = L\{-2x\} \qquad \end{aligned}

\begin{aligned} s^2X - sx(0) - \dot{x(0)} = -2X \qquad \end{aligned}

Now we have to solve for X and use our initial conditions.

\begin{aligned} s^2X - s*3 - 4 = -2X \qquad \end{aligned}

\begin{aligned} s^2X + 2X = 3s + 4 \qquad \end{aligned}

\begin{aligned} X * (s^2 + 2) = 3s + 4 \qquad \end{aligned}

\begin{aligned} X = \frac{3s + 4}{s^2 + 2} \qquad \end{aligned}

Now we have several options:

a/ Do partial fraction decomposition by hand - described here and here and take inverse Laplace transformation.

b/ Do partial fraction decomposition in Matlab/Octave - use command residue and take inverse Laplace transformation.

c/ Use online Laplace calculator (here or here) to compute inverse Laplace transformation of our term. In our case it would be - link.

Either way, the solution we get should be:

\begin{aligned} L^{-1}\{X\} = 3\cos \left(\sqrt{2}t\right)+\frac{4\sin \left(\sqrt{2}t\right)}{\sqrt{2}} \qquad \end{aligned}

## Comments

watch free exope faste for anytime and 48.

dirty lesbo sluts, japanese alternative porn, [censored] her tight ass, nudes beatiful mature, free porn downloads dutch, youtube mature video.

see all the not safe for work pics, gifs and most popular animated gifs here on giphy. find funny gifs, cute gifs, reaction gifs and more.

free penelope cruz porn videos, bbw slut [censored]s, sluts on the loose, truth or dare blowjob, free skinny lesbians, lita getting [censored]ed, what the [censored] is a dreadnought.

c2231c2074322.com,

c2231c2074322.com,

http://c2231c2074322.com

in news papers but now as I am a user of net thus from now

I am using net for articles, thanks to web.

okrem toho som spokojný, lebo konečne niekto píše stručne a k veci.

Budem sledovať stránku

RSS feed for comments to this post