# R2.1

Given the two roots and the initial conditions:

 $\displaystyle \lambda_1=-2, \lambda_2=+5$ $\displaystyle (Eq. 1a)$
 $\displaystyle y(0)=1,y'(0)=0$ $\displaystyle (Eq. 1b)$

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $\displaystyle r(x)$

Consider no excitation:

 $\displaystyle r(x)=0$ $\displaystyle (Eq. 2)$

Plot the solution.

Characteristic equation:

 $\displaystyle (\lambda-\lambda_1)(\lambda-\lambda_2)=0$ $\displaystyle (Eq. 3)$

Substituting $\displaystyle(Eq. 1a)$ into $\displaystyle(Eq. 3)$:

 $\displaystyle (\lambda+2)(\lambda-5)=0$ $\displaystyle (Eq. 4)$

 $\displaystyle \lambda^2-3\lambda-10=0$ $\displaystyle (Eq. 5)$

Non-homogeneous solution:

 $\displaystyle y''-3y'-10y=0$ $\displaystyle (Eq. 6)$

Homogeneous solution:

 $\displaystyle y_h(x)=C_1e^{-2x}+C_2e^{5x}$ $\displaystyle (Eq. 7)$

Overall solution:

 $\displaystyle y(x)=C_1e^{-2x}+C_2e^{5x}+y_p(x)$ $\displaystyle (Eq. 8)$

No excitation:

 $\displaystyle r(x)=0 => y_p(x)=0$ $\displaystyle (Eq. 9)$

From intitial conditions:

 $\displaystyle y(0)=1=C_1+C_2$ $\displaystyle (Eq. 10)$

 $\displaystyle y'(0)=0=-2C_1+5C_2$ $\displaystyle (Eq. 11)$

Solving for $\displaystyle C_1$ and $\displaystyle C_2$:

 $\displaystyle C_1=\frac{5}{7}$ $\displaystyle$

and

 $\displaystyle C_2=\frac{2}{7}$ $\displaystyle$

So, the final solution is:

$\displaystyle y(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x}$

# R2.2

 $\displaystyle y'' - 10y' +25y = r(x)$ $\displaystyle (Eq. 1)$

Initial conditions:

 $\displaystyle y(0) = 1, y'(0) = 0$ $\displaystyle (Eq. 2)$

No excitation:

 $\displaystyle r(x) = 0$ $\displaystyle (Eq. 3)$

Find and plot the solution for $\displaystyle (Eq. 1)$

Due to no excitation, $\displaystyle (Eq. 1)$ becomes:

 $\displaystyle y'' - 10y' +25y = 0$ $\displaystyle (Eq. 4)$

Substituting $\displaystyle \lambda$ into $\displaystyle (Eq. 4)$:

 $\displaystyle \lambda^2-10\lambda+25=0$ $\displaystyle (Eq. 5)$

Factoring, and solving for $\displaystyle \lambda$:

 $\displaystyle (\lambda-5)^2=0$ $\displaystyle (Eq. 6)$

 $\displaystyle \lambda_2=\lambda_1=\lambda=5$ $\displaystyle$

Since $\displaystyle \lambda$ is a double root, the general solution:

 $\displaystyle y=C_1e^{5x}+C_2xe^{5x}$ $\displaystyle (Eq. 7)$

From intitial conditions:

 $\displaystyle y(0)=1=C_1$ $\displaystyle (Eq. 10)$

 $\displaystyle y'(0)=0=5C_1+C_2$ $\displaystyle (Eq. 11)$

Solving for $\displaystyle C_1$ and $\displaystyle C_2$:

 $\displaystyle C_1=1$ $\displaystyle$

and

 $\displaystyle C_2=-5$ $\displaystyle$

So, the final solution is:

$\displaystyle y(x)=e^{5x}-5xe^{5x}$

# R1.6G

G) Deformation of a Beam

$\displaystyle{EIy''=f(x)}$

$\cdot$ This equation is of SECOND order and LINEAR.

$\cdot$ To prove the applicability of Superposition, the following is done:

Homogeneous

 $\displaystyle{EIy_h''=0}$ $\displaystyle (Eq. 1)$

Particular:

 $\displaystyle{EIy_p''=f(x)}$ $\displaystyle (Eq. 2)$

Add (Eq. 1) and (Eq. 2):

$\displaystyle{EIy_h''+EIy_p''=f(x}$

It could be said that:

$\displaystyle{y_{h}+y_{p}=Y}$

and due to the fact that,

$\displaystyle{EI(y_h''+y_p)=EIY}$

 It can be said that Y is a solution, confirming that superposition CAN be used.

# R1.1

Team14P1R16Diagram1.png‎

File:Team14P1R11Diagram2.png
Free Body Diagram

$y=y_k=y_c$

Spring Force, $\vec{F}_s=k(y)$

Dashpot Force, $\vec{F}_D=c(y')$

Applied Force, $\vec{F}(t)$

 $\displaystyle \Sigma\vec{F}=\vec{F}(t)-\vec{F}_s-\vec{F}_D$ $\displaystyle (Eq. 1)$

 $\displaystyle m\vec{a}=\vec{F}_i=my''$ $\displaystyle (Eq. 2)$

From Newton's 2nd Law:

 $\displaystyle \Sigma\vec{F}=m\vec{a}$ $\displaystyle (Eq. 3)$

Substituting $(Eq. 1)$ and $(Eq. 2)$ into $(Eq. 3)$:

 $\displaystyle \vec{F}(t)-\vec{F}_s-\vec{F}_D=\vec{F}_i$ $\displaystyle (Eq. 4)$

Solving for $\vec{F}(t)$:

$\vec{F}(t)=\vec{F}_s+\vec{F}_D+\vec{F}_i$

Substituting $\vec{F}_s=ky$, $\vec{F}_D=cy'$, and $\vec{F}_i=my''$ into $(Eq. 4)$:

$\vec{F}(t)=ky+Cy'+my''$

Putting into standard form:

$my''+Cy'+ky=\vec{F}(t)$