User:Egm4313.s12.team2.sleshko

About Me
Hey team. My name is Stephen.

--Egm4313.s12.team2.sleshko 21:39, 22 January 2012 (UTC)

=Report 1=

Problem Statement
Derive the equation of motion of the spring-mass-dashpot in figure 53. on page 85 of Kreyszig's Advanced Engineering Mathematics (2011).

Kinematics
$$ \bar{r} = y \bar{E}_y $$ $$ \bar{v}=y' \bar{E_y}$$ $$ \bar{a} = \bar{v}' \bar{E_y}=y''\bar{E_y}$$

Description of Kinematics
The kinematics describing the motion of the spring-dashpot-mass system in Fig. 53 (also pictured above) are derived, first, from the position function of the mass with respect to time $$\bar{r}(t) = y\bar{E}_y $$. The next expression, $$\bar{v}={y}'\bar{E}_y$$,is the function for velocity, which is the derivative of the position function $$\bar{r}(t) = y\bar{E}_y $$. The expression, $$\bar{a}=y''\bar{E}_y$$ represents the acceleration of the mass with respect to time, and is the derivative of the velocity function, $$\bar{v}={y}'\bar{E}_y$$.

From Newton's 2nd Law of Motion
$$\bar{F} = m\bar{a} $$ $$\bar{F}_s = -ky\bar{E}_y$$ $$\bar{F}_d = cy'\bar{E}_y $$ $$ mg = m\bar{a} = my''\bar{E}_y$$ Where $$\bar{F}_s$$ is equal to the force in the spring and $$\bar{F}_d$$ is equal to the force in the dashpot
 * Sum the Forces

$$ ky\bar{E}_y + cy'\bar{E}_y + my''\bar{E}_y = r(t) $$


 * From Kreyszig pg. 86

$$ \bar{r}(t) = \bar{F}_0{cos}(wt) $$ $$ ky\bar{E}_y + cy'\bar{E}_y + my''\bar{E}_y = \bar{F}_0{cos}(wt) $$ --Egm4313.s12.team2.sleshko 20:43, 27 January 2012 (UTC)

Problem (5)
$$ y'' +2\pi y' +\pi y = 0 \ $$ $$ r^2 +2\pi r +\pi ^2 = 0 \ $$ $$r = \frac{-2\pi \pm \sqrt{4\pi ^2 - 4\pi^2} }{2} \ $$ $$ r= \frac{-2\pi}{2} = -\pi\ $$ Therefore, the general solution takes the form of:

$$ y_g = C_1e^{-\pi t} + C_2te^{-\pi t} \ $$ $$ y' = -\pi C_1e^{-\pi t} + C_2[e^{-\pi t} -\pi te^{-\pi t}] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi e^{-\pi t} - \pi(e^{-\pi t}-\pi te^{-\pi t})] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi ^2te^{-\pi t} - 2\pi e^{-\pi t}] \ $$

Substitution back into the original equation should result as 0=0

$$0 = \pi^2C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} -2\pi C_2e^{-\pi t} - 2\pi^2C_1e^{-\pi t} + 2\pi C_2e^{-\pi t} -2\pi^2C_2e^{-\pi t}+\pi^2 C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} \ $$ $$ 0 = 0 \ $$ Thus the substitution checks out, and the form of the general solution is proven.

--Stephen Leshko 20:10, 30 January 2012 (UTC)

=Report 2=

Problem (5)
$$ y'' +2\pi y' +\pi y = 0 \ $$ $$ r^2 +2\pi r +\pi ^2 = 0 \ $$ $$r = \frac{-2\pi \pm \sqrt{4\pi ^2 - 4\pi^2} }{2} \ $$ $$ r= \frac{-2\pi}{2} = -\pi\ $$ Therefore, the general solution takes the form of:

$$ y_g = C_1e^{-\pi t} + C_2te^{-\pi t} \ $$ $$ y' = -\pi C_1e^{-\pi t} + C_2[e^{-\pi t} -\pi te^{-\pi t}] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi e^{-\pi t} - \pi(e^{-\pi t}-\pi te^{-\pi t})] \ $$ $$ y'' = \pi ^2 C_1e^{-\pi t} + C_2[-\pi ^2te^{-\pi t} - 2\pi e^{-\pi t}] \ $$

Substitution back into the original equation should result as 0=0

$$0 = \pi^2C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} -2\pi C_2e^{-\pi t} - 2\pi^2C_1e^{-\pi t} + 2\pi C_2e^{-\pi t} -2\pi^2C_2e^{-\pi t}+\pi^2 C_1e^{-\pi t} + \pi^2C_2te^{-\pi t} \ $$ $$ 0 = 0 \ $$ Thus the substitution checks out, and the form of the general solution is proven.

--Stephen Leshko 20:10, 30 January 2012 (UTC)

Problem 6

 * $$ 10y'' -32y' +25.6y = 0 \ $$

Which can be written in standard form as:
 * $$ y'' -3.2y' +2.56y = 0 \ $$

The characteristic equation is then:
 * $$ \lambda ^2 - 3.2\lambda +2.56 = 0 \ $$
 * $$ \lambda_1,\lambda_2 = \frac{3.2 \pm \sqrt{(-3.2)^2 - 4(2.56)}}{2} \ $$

Therefore the roots $$ \lambda_1, \lambda_2 = 1.6 \ $$ are actually a double root at $$ 1.6 $$ The form for the solution of the differential equation with double-real roots is of the following:
 * $$ y_h(t) = C_1e^{1.6t} + C_2te^{1.6t} \ $$

And then in order to check the solution, find the first and second derivatives to plug back into the original equation and check their equality to zero.
 * $$ y'_h(t) = 1.6C_1 e^{1.6t} + C_2[e^{1.6t} +1.6te^{1.6t}] \ $$
 * $$ y''_h(t) = 2.56C_1e^{1.6t} + C_2[1.6e^{1.6t} + 1.6e^{1.6t} + 2.56te^{1.6t}] \ $$


 * $$ 10[1.6C_1e^{1.6t} +C_2[3.2e^{1.6t} +2.56te^{1.6t}]] - 32[1.6C_1e^{1.6t} +C_2[e^{1.6t} + 1.6te^{1.6t}]] + 25.6[C_1e^{1.6t} +C_2te^{1.6t}] = 0 \ $$


 * $$ 25.6C_1e^{1.6t} + 32C_2e^{1.6t} + 25.6C_2te^{1.6t} - 51.2C_1e^{1.6t} -32C_2e^{1.6t} - 51.2C_2te^{1.6t} + 25.6C_1e^{1.6t} + 25.6C_2te^{1.6t} = 0 \ $$

The solution then checks out

Problem Statement
Find an ODE $$ y'' +ay' + by = 0 \ $$ for the given basis.

Problem (16)
Basis: $$ e^{2.6x}, e^{-4.3x} \ $$


 * $$ \lambda_1,\lambda_2 =2.6, -4.3 = \frac{-a \pm \sqrt{a^2 - 4b}}{2} \ $$

Therefore multiplying each $$ \lambda \ $$ by 2 yields:
 * $$ 5.2 = -a + \sqrt{a^2 - 4b} \ $$ ............. (1)
 * $$ -8.6 = -a - \sqrt{a^2 -4b} \ $$ ............. (2)

Adding the equations together gives:
 * $$ -3.4 = -2a \ $$ so $$ a = 1.7 \ $$

Plug $$ a \ $$ back into equation (1) to get:
 * $$ 5.2 = -1.7 + \sqrt{1.7^2 - 4b} \ $$

Solving for $$ b \ $$:
 * $$ b = \frac {(5.2 +1.7)^2 - 1.7^2}{-4} = -11.18 \ $$

So then the differential equation would be:
 * $$ y'' +1.7y' - 11.18 y = 0 \ $$

Problem (17)
Basis: $$ e^{-\sqrt{5}x} ,xe^{-\sqrt{5}x} \ $$

From the basis it can be seen that $$ -\sqrt{5} \ $$ is a double root of the characteristic equation
 * $$ \lambda_1, \lambda_2 = - \sqrt{5} \ $$

So then the characteristic equation must have root such that:
 * $$ (\lambda + \sqrt{5})^2 = 0 \ $$

Multiplying out gives:
 * $$ \lambda ^2 + 2 \sqrt{5} \lambda + 5 = 0 \ $$

Which corresponds to a differential the equation:
 * $$ y'' + \sqrt{20} y' + 5y = 0 \ $$