User:Egm6321.f10.team6.lee/HW5

=Equation of Motion of a Spring-Damper-Mass System etc…=

Problem Statement
A general spring-mass-damper system has spring constant k, mass m, and damping coefficient c. The equation of motion is given as :

Part 1
1. Find the PDE's integration factor h(t,y). (Condition 2 of exactness - 2 relations)

Part 2
let the trial solution for integrating factor $$ \begin{align} h(t)={e}^{\alpha t} \end{align}$$

Multiply the integrating factor to the eqn(5.1) and integrate both side.

Assume that the left hand side would have the form of eqn(5.3) to reduce order.

1. Find $$ \begin{align} \bar{{a}_{1}} \end{align}$$ and $$ \begin{align} \bar{{a}_{2}} \end{align}$$ in terms of $$ \begin{align} {a}_{2},{a}_{1},{a}_{0} \end{align}$$

2. Find quadratic equation for $$\begin{align} \alpha \end{align} $$.

3. Reduce order of equation.

4. Find $$\begin{align} y(t) \end{align} $$ by applying integrating factor.

5. Show that:

Where $$\begin{align} \alpha \end{align} $$ and $$\begin{align} \beta \end{align} $$ are roots of the equation for $$\begin{align} \alpha \end{align} $$ in part 2.2

6. Deduce and expression for the particular solution $$\begin{align} {y}_{P} \end{align} $$for a general excitation  $$\begin{align} f(t) \end{align}$$

7. Verify with the table of particular solutions. Consider, $$\begin{align} f(t)={t}^{2} \end{align} $$

8. Solvev L2_ODE_CC when $$ f(t)= {e}^{-{t}^{2}}$$ and find coefficient $$\begin{align} {a}_{2},{a}_{1},{a}_{0} \end{align} $$ :

such that

8.1 The roots for equation 5.1 are obtained from $$\left( r+1 \right)\left( r-2 \right)=0$$.

8.2 The roots for equation 5.1 are obtained from $${{\left( r-4 \right)}^{2}}=0$$. .

Part 1.1
Let's find the integrating factor $$\begin{align} h(t,y) \end{align}$$ first.

Multiply the integrating factor to the eqn(5.1).

and we can rewrite the eqn(5.6) to check the exactness condition.

where $$\begin{align} p:= y' \end{align} $$

therefore,

$$\begin{align} f(x,y,p) = h(t,y){a}_{2} \end{align}$$

$$\begin{align} g(x,y,p) = h(t,y){{a}_{1}}p+ h(t,y) {{a}_{0}}y - h(t,y)\cdot f(t)\end{align}$$

There are two relations for the 2nd condition of exactness.

we can check the second relation i.e eqn(5.9) first as it is less complex than eqn(5.8).
 * The second condition of exactness

Hence,

Let's check the eqn(5.8). we can drop out $$\begin{align} {h}_{y}, {h}_{ty}, {h}_{yy} \end{align}$$ for the eqn(5.11)

and then, we get the L2_ODE_CC. i.e.

Hence, we can find $$\begin{align} h(t,y) \end{align}$$ from the eqn(5.14).

Part 2.1
As eqn(5.3) is assumed to be the same to eqn(5.2),

Differentiate both side of the eqn(5.15) i.e.

therefore,

Part 2.2
We can find the quadric equation for $$\begin{align} \alpha \end{align}$$ from the eqn(5.18),

Multiply $$\begin{align} \alpha \end{align}$$ both side.

Part 2.3
Let's reduce order for the eqn(5.2).

As we assumed that the eqn(5.2) and eqn(5.3) are the same, we can put them as,

Part 2.4
by using integrating factor method [eqn(1)p10-1 at the lecture note] where, $$\begin{align} \beta := \frac{\bar{{a}_{0}}} \end{align}$$

Let's find $$\begin{align} y(t) \end{align}$$ [eqn(6)p10-3 at the lecture note]

Part 2.5
We know that

Therefore,

As eqn(5.28) is satisfied, we can set quadric equation which has two roots $$\begin{align} \alpha, \beta\end{align}$$

This is the same as eqn(5.20).

Part 2.6
Let's rearrange the eqn(5.26) to distinguish $$\begin{align} {y}_{H1}, {y}_{H2}, {y}_{p} \end{align}$$

Part 2.7
Let's check the eqn(5.32) with the table of particular solution when $$ \begin{align}f(t)={t}^{2}\end{align}$$

Assume that, then,
 * with the table of particular solution

we can put them to the eqn(5.1).

there are three unknowns and theree equations. therefore we can find A, B and C. therefore, the particular solution by using the table would be,

this eqn(5.36) should be the same to the particular solution with eqn(5.26).

from eqn(5.32), we can integrate this equation easily with the help of Wolfram Integrator we can put the eqn(5.27) and eqn(5.28) to the eqn(5.37).
 * with general particular solution - eqn(5.32)

Now, we have verified that the particular solution (5.32) is correct because eqn(5.38) is the same to the eqn(5.36)

2.8.1
from given characteristic equation, $$ \begin{align} (r+1)(r-2)=0 \end{align}$$

we can find two homogeneous solutions,as $$ \begin{align} {r}_{1}=-1, {r}_{2}= 2 \end{align}$$.

from eqn(5.31), we know that

therefore,

we can find the coefficients $$ \begin{align} {a}_{0}, {a}_{1}, {a}_{2} \end{align}$$. from the eqn(5.27) & (5.28).

The homogeneous differential equation is,

we can check the result with WolframAlpha, and it shows that there are two homogeneous solutions which are the same to eqn(5.40)

Now, let's apply the Gaussian distribution to the f(t). then,

By using the WolframAlpha, we can find y(t). ♦see the result

2.8.2
from given characteristic equation, $$ \begin{align} {(r-4)}^{2}=0 \end{align}$$

we can find two homogeneous solutions,as $$ \begin{align} {r}_{1}={r}_{2}= 4 \end{align}$$.

(they are the same double roots.)

from eqn(5.31), we know that

therefore,

we can find the coefficients $$ \begin{align} {a}_{0}, {a}_{1}, {a}_{2} \end{align}$$. from the eqn(5.27) & (5.28).

The homogeneous differential equation is,

we can check the result with WolframAlpha, and it shows that there are two homogeneous solutions which are the same to eqn(5.45)

Now, let's apply the Gaussian distribution to the f(t). then,

By using the WolframAlpha, we can find y(t). ♦see the result