User:Egm4313.s12.team3.droll

=Problem 4: Find a General Solution=

Problem Statement
For the following two problems find the general solution for the equations and then check the answer by substitution.
 * {| style="width:100%" border="0"

$$\displaystyle y''+2\pi y'+\pi^2 y=0 $$ $$\displaystyle 10y''-32y'+25.6y=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Part A
This problem is very similar to [| EGM4313 Team 3's Report 1 Problem 5, Part B]
 * {| style="width:100%" border="0"

$$\displaystyle y''+2\pi y'+\pi^2 y=0 $$ (4.1) From the characteristic equation (found on K 2011 p.54 (3)):
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+a\lambda+b=0 $$ (4.2) Using (4.2) and substituting the corresponding values for a and b you get:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+2\pi \lambda+\pi^2=0 $$ (4.3) Next we will check for the possibility of a Real Double Root (K 2011 p.55 "Case II") using:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a^2-4b=0 $$ (4.4)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle (2\pi)^2-4(\pi^2)=0 $$ $$  \displaystyle 4\pi^2 - 4\pi^2=0 $$ $$  \displaystyle 0=0 $$ Since the Equation is a Real Double Root you now use:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y=(C_1+C_2x)e^{-ax/2} $$ (4.5) Solving for $$\displaystyle -a/2 $$ equals:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{-a}{2}=\frac{-2\pi}{2}=-\pi $$ (4.6) Substituting (4.6) into (4.5) leaves:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y=(C_1+C_2x)e^{-\pi x} $$ (4.7) Taking the first and second derivative of (4.7) gives:
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y'=-\pi C_1e^{-\pi x}+C_2e^{-\pi x}-\pi C_2xe^{-\pi x} $$ (4.8)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y''=\pi^2 C_1e^{-\pi x}-\pi C_2e^{-\pi x}+\pi^2 C_2xe^{-\pi x}-\pi C_2e^{-\pi x} $$ (4.9) Finally inserting (4.7),(4.8), and (4.9) into (4.1) allows a check of the answer:
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle (\pi^2 C_1e^{-\pi x}-\pi C_2e^{-\pi x}+\pi^2 C_2xe^{-\pi x}-\pi C_2e^{-\pi x})+2\pi (-\pi C_1e^{-\pi x}+C_2e^{-\pi x}-\pi C_2xe^{-\pi x})+\pi^2 ((C_1+C_2x)e^{-\pi x})=0 $$ (4.10)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Through simple canceling of like terms from (4.10) gives:
 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Part B
Going over many of the same steps in Part A we have the following equation:
 * {| style="width:100%" border="0"

$$\displaystyle 10y''-32y'+25.6=0 $$ (4.11) Which when divided by 10 equals:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y''-3.2y'+2.56=0 $$ (4.12) From the characteristic equation (again found on K 2011 p.54 (3)):
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+a\lambda+b=0 $$ (4.13) Using (4.12) and substituting the corresponding values for a and b you get:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2-3.2 \lambda+2.56=0 $$ (4.14) We will again check for the possibility of a Real Double Root (also on K 2011 p.55 "Case II") using:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle a^2-4b=0 $$ (4.15)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle (-3.2)^2-4(2.56)=0 $$ $$  \displaystyle 10.24 - 10.24=0 $$ $$  \displaystyle 0=0 $$ Since this equation is also a Real Double Root you use:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y=(C_1+C_2x)e^{-ax/2} $$ (4.16) Solving for $$\displaystyle -a/2 $$ equals:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{-a}{2}=\frac{-(-3.2)}{2}=1.6 $$ (4.17) Substituting (4.17) into (4.16) gives the equation:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y=(C_1+C_2x)e^{1.6x} $$ (4.18) Taking the first and second derivative of (4.18) gives:
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y'=((1.6)(C_1+C_2x)+C_2)e^{1.6x} $$ (4.19)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }
 * {| style="width:15%" border="0"

$$  \displaystyle y''=2.56C_1e^{1.6x}+((2+1.6x)(1.6C_2e^{1.6x})) $$ (4.20) BY inserting (4.18),(4.19), and (4.20) into (4.11) the answer can be verified:
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle 10(2.56C_1e^{1.6x}+((2+1.6x)(1.6C_2e^{1.6x})))-32(((1.6)(C_1+C_2x)+C_2)e^{1.6x})+25.6((C_1+C_2x)e^{1.6x})=0 $$ (4.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

As in Part A simple canceling of similar terms gives:
 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * }

=Problem 3: Spring-Dashpot-Mass System in Series=

Problem Statement
For the following system (as can be seen in the figure below) comprised of a spring, a dashpot, and a mass in series solve for the equation of motion and draw the free body diagrams.

Problem Solution
For information regarding the order and linearity of ordinary differential equations, see Loc Vu-Quoc class EGM 4313 notes Sec 1. Also, from Sec 1 of Loc Vu-Quoc's class EGM4313 (p.1-4).

Kinematics

 * {| style="width:100%" border="0"

$$\displaystyle {y}=y_k+y_c $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"


 * <p style="text-align:right">
 * <p style="text-align:right">
 * }

Where:
 * {| style="width:100%" border="0"

$$\displaystyle {y_k}=Spring\,Displacement $$
 * style="width:95%" |
 * style="width:95%" |
 * }

&
 * {| style="width:100%" border="0"

$$\displaystyle {y_c}=Dashpot\,Displacement $$
 * style="width:95%" |
 * style="width:95%" |
 * }

Kinetics

 * {| style="width:100%" border="0"

$$\displaystyle {a}=d^2y/dt^2={y}'' $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From th previous equation for acceleration and the mass (represented by the variable m) the sum of the forces can represented as follows:
 * {| style="width:100%" border="0"

$$\displaystyle m{y}''=f(t)-f_I $$     (2) Where:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle f_I=The\,Internal\,Forces\,on\,the\,Mass $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Taking the second derivative of (1):
 * {| style="width:100%" border="0"

$$\displaystyle {(y=y_k+y_c)}'' $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Equals
 * {| style="width:100%" border="0"

$$\displaystyle {y}={y_k}+{y_c}'' $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From the free body diagrams above through geometry :
 * {| style="width:100%" border="0"

$$\displaystyle f_I=f_k $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

And:
 * {| style="width:100%" border="0"

$$\displaystyle f_k=f_c $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Which finally means that:
 * {| style="width:100%" border="0"

$$\displaystyle f_I=f_c=f_k $$     (7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

With:
 * {| style="width:100%" border="0"

$$\displaystyle f_k=k{y_k} $$     (8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

&
 * {| style="width:100%" border="0"

$$\displaystyle f_c=C{y_c}' $$     (9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

From (7) we can solve for $$\displaystyle {y_c}'$$
 * {| style="width:100%" border="0"

$$\displaystyle k{y_k}=C{y_c}' $$     (10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$\displaystyle {y_c}'=(k/C){y_k} $$     (11) Then solving for $$\displaystyle {y_c}''$$ equals:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$\displaystyle {(y_c')}'=((k/C)y_k)'\,\,==>{y_c}''=(k/C){y_k}' $$     (12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Then plugging (12) into (4):
 * {| style="width:100%" border="0"

$$\displaystyle {y}=y_k+(k/C){y_k}' $$     (13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substituting (13) for $$\displaystyle {y}''$$
 * {| style="width:100%" border="0"

$$\displaystyle m({y_k''+(k/C){y_k}'})=f(t)-f_I $$     (14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

And solving for $$\displaystyle f(t)$$ with $$\displaystyle f_I=$$ (5)
 * {| style="width:100%" border="0"

$$\displaystyle m({y_k''+(k/C){y_k}'})+f_k=f(t) $$     (15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Finally with (8) substituted into $$\displaystyle f_k$$ in (15) gives the final answer of:
 * {| style="width:100%" border="0"

$$\displaystyle m({y_k''+(k/C){y_k}'})+k{y_k}=f(t) $$     (16)
 * style="width:95%" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }