User:Egm4313.s12.team4/Report 1

= Problem 1: Equation of Motion of a Spring-Dashpot System =

Problem Statement
Derive the equation of motion of a spring-dashpot system in parallel with a mass and applied force, $$\displaystyle f(t)$$



Solution
Using a force balance, the following equation can be derived:


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

$$\displaystyle f(t) = my'' + f_I$$ Where $$\displaystyle f_I = f_c = f_k$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.1)
 * }

Adding the displacements due to the spring $$\displaystyle y_k$$ and the dashpot $$\displaystyle y_c$$ we can write the following equation:


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

$$\displaystyle y=y_k+y_c$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.2)
 * }

Taking two time derivatives of (1.1.2) gives:


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$$\displaystyle y = y_k + y''_c$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.3)
 * }

Since $$\displaystyle f_k=f_c$$, we can write:


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$$\displaystyle ky_k = cy'_c$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.4)
 * }

Now, solving for $$\displaystyle y'_c $$ in (1.1.4), we find:


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$$\displaystyle y'_c = \frac{k}{c}y_k$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.5)
 * }

We can substitute our new value for $$\displaystyle y'_c$$ into (1.1.3) to obtain:


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

$$\displaystyle y = y_k + (y'_c)'$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.6)
 * }


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$$\displaystyle y = y_k + (\frac{k}{c}y_k)' $$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.7)
 * }


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

$$\displaystyle y = y_k + \frac{k}{c}y'_k$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.1.8)
 * }

We can now substitute for $$\displaystyle y''$$ and $$\displaystyle f(I)$$ in equation (1.1.1) to obtain an equation in terms of $$\displaystyle y_k$$:


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


 * style="width:95%" |
 * style="width:95%" |
 * (1.1.9)
 * }

= Problem 2: Equation of Motion of a Spring-Mass-Dashpot System =

Problem Statement
Derive the equation of motion of the spring-mass-dashpot in Fig. 53, in K2011 pg. 85 with an applied force r(t) on the ball.



Solution
In order to derive an equation of motion, all of the forces must be analyzed. In the spring-mass-dashpot model there are forces from the spring, inertia, and a damping force caused by the dashpot.

The force of a spring oriented in the y-direction is given by Hooke's Law:
 * {| style="width:100%" border="0"

$$\displaystyle F=-ky$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.1)
 * }

The damping force of a dashpot is proportional to the mass' velocity. Relating velocity to position and time:
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$$\displaystyle V=\frac{dy}{dt}$$
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 * style="width:95%" |
 * (1.2.2)
 * }

Proportionality is related by a dashpot coefficient $$\displaystyle c$$
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$$\displaystyle F_{d}=c\frac{dy}{dt}$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.3)
 * }

According to Newton's Second Law of Motion
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$$\displaystyle \Sigma F_{y}=ma_{y}$$
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 * style="width:95%" |
 * (1.2.4)
 * }

Relating acceleration to position and time
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$$\displaystyle a_{y}=\frac{\mathrm{d^2}y}{\mathrm{d} t^2}$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.5)
 * }

Summing spring and damping forces and substituting new acceleration term
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$$\displaystyle m\frac{\mathrm{d^2}y}{\mathrm{d} t^2}=-c\frac{dy}{dt}-ky$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.6)
 * }

Both forces are negative because they resist the motion of the spring.

Moving all terms to one side
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$$\displaystyle m\frac{\mathrm{d^2}y}{\mathrm{d} t^2}+c\frac{dy}{dt}+ky=0$$
 * style="width:95%" |
 * style="width:95%" |
 * (1.2.7)
 * }

Setting the equation equal to $$\displaystyle r(t)$$
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.2.8)
 * }

=Problem 3: FBDs and Equation of Motion of a Spring-Dashpot System =

Problem Statement
For the spring-dashpot-mass system on p.1-4, draw the free body diagrams and derive the equation of motion (2) p.1-4.



Solution
Free body diagram:

Force balance equation of spring-dashpot-mass system:
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$$\displaystyle \sum F_x=f(t)-f_I=ma$$ where $$\displaystyle a={y}''\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.1)
 * }

Rearranging
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$$\displaystyle m{y}''+f_I=f(t)\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.2)
 * }

Internal force
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$$\displaystyle f_I=f_k=f_c\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.3)
 * }

Force relationships
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$$\displaystyle f_k=ky_k\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.4)
 * }


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$$\displaystyle f_c={y_c}''\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.5)
 * }

Where

$$\displaystyle c = $$ dashpot constant

$$\displaystyle k = $$ spring constant

Total displacement is displacement from the spring plus displacement from the dashpot:
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$$\displaystyle y=y_k+y_c\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.6)
 * }

Taking the double integral of $$\displaystyle y$$ you get
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$$\displaystyle {y}={y_k}+{y_c}''\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.7)
 * }

Get $$\displaystyle y_c$$ in terms of $$\displaystyle y_k$$:
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$$\displaystyle f_k=f_c\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.8)
 * }


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$$\displaystyle ky_k=c{y_c}'\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.9)
 * }


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$$\displaystyle {y_c}''=\frac{k}{c}y_k\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.10)
 * }

Plug back into $$\displaystyle {y}''\,\!$$:
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$$\displaystyle {y}={y_k}+{({y_c}')}'\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.11)
 * }


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$$\displaystyle {({y_c}')}'=\frac{k}{c}{y_k}'\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.12)
 * }


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$$\displaystyle \rightarrow {y}={y_k}+\frac{k}{c}{y_k}'\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.13)
 * }

Now get $$\displaystyle f_I$$ in terms of $$\displaystyle y_k$$:
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$$\displaystyle f_I=f_k\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.14)
 * }


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$$\displaystyle f_k=ky_k\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.15)
 * }


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$$\displaystyle \rightarrow f_I=ky_k\,\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.16)
 * }

When plugging in all variables into original equation you get:
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 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.3.17)
 * }

= Problem 4: Deriving Capacitance Equations =

Problem Statement
Derive equations (1.4.2) and (1.4.3) from (1.4.1).


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$$\displaystyle V=LC \frac{d^2 v_C}{dt^2} + RC \frac{d v_C}{dt} + v_C$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.1)
 * }


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$$\displaystyle L I'' + R I' + \frac{1}{C} I = V'$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.2)
 * }


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$$\displaystyle L Q'' + R Q' + \frac{1}{C} Q = V$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.3)
 * }

Background Information

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$$\displaystyle C = \frac{Q}{V}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.4)
 * }


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$$\displaystyle I = C \frac{d v_C}{d t}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.5)
 * }

Part A
Taking the derivative of (1.4.1)
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$$\displaystyle V' = LC \frac{d^3 v_C}{d t^3} + RC \frac{d^2 v_C}{d t} + \frac{d v_C}{d t}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.6)
 * }

Rewrite equation (1.4.5)
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$$\displaystyle I^{(n-1)} = C \frac{d^n v_C}{d t^n}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.7)
 * }

Substituting (1.4.7) into (1.4.6)
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$$\displaystyle V' = L I^{(2)} + R I^{(1)} + \frac{1}{C} I^{(0)}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.8)
 * }

Rewrite
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 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.2)
 * }

Part B
Rewrite equation (1.4.4)
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$$\displaystyle Q^{(n)} = C V^{(n)}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.9)
 * }

Substituting (1.4.9) into (1.4.1)
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$$\displaystyle V = L Q^{(2)} + R Q^{(1)} + \frac{1}{C} Q^{(0)}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.10)
 * }

Rewrite
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 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.4.3)
 * }

= Problem 5a: General Solution for ODEs =

Problem Statement
Problem found in the textbook on page 59, problem 4. Find a general solution. Check your answer by substitution.


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$$\displaystyle y'' + 4y' + (\pi^2 +4)y = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.1)
 * }

Solution
The equation above has a characteristic equation:
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$$\displaystyle \lambda^2 + 4\lambda + (\pi^2 +4) = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.2)
 * }

which is of the form:
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$$\displaystyle \lambda^2 + a\lambda + b = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.3)
 * }

because the discriminant
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$$\displaystyle a^2 - 4b < 0$$ making it negative.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.4)
 * }

The roots of the characteristic equation are:
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$$\displaystyle \lambda = -\frac{1}{2}a\pm wi $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.5)
 * }

where
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$$\displaystyle w^2 = b - \frac{1}{4}a^2$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.6)
 * }

Therefore real solutions can be written as:
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$$\displaystyle y_1 = e^{-\frac{ax}{2}}\cos wx$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.7)
 * }

and
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$$\displaystyle y_2 = e^{-\frac{ax}{2}}\sin wx$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.8)
 * }

This will yield a general solution of:
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$$\displaystyle y = e^{-\frac{ax}{2}}(A \cos wx + B \sin wx)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.9)
 * }

where A and B are arbitrary.

Solving for $$\displaystyle w$$, the roots become:
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$$\displaystyle \lambda = -2 \pm \pi i $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.10)
 * }

The final general solution for this problem is:
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.11)
 * }

Check by substitution:

In order to check the solution we must find $$\displaystyle y' $$ and $$\displaystyle y'' $$ :

Now that we have
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$$\displaystyle y= e^{-2x}(A \cos \pi x + B \sin \pi x)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.12)
 * }


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

$$\displaystyle y'= -2e^{-2x}(A \cos \pi x + B \sin \pi x) + e^{-2x}(-A \pi \sin \pi x + B \pi \cos \pi x)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.13)
 * }

and
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$$\displaystyle y''= 4e^{-2x}(A \cos \pi x + B \sin \pi x) - 2e^{-2x}(-A \pi \sin \pi x + B \pi \cos \pi x)$$ $$\color{White}y''= \color{Black} - 2e^{-2x}(-A \pi \sin \pi x + B \pi \cos \pi x) + e^{-2x}(-A \pi ^2 \cos \pi x - B \pi ^2 \sin \pi x)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.14)
 * }

we can substitute these equations into the original equation,
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$$\displaystyle y'' + 4y' + (\pi^2 +4)y = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.15)
 * }

After substitution we see that
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$$\displaystyle 0 = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5a.16)
 * }

Therefore the solution is correct.

= Problem 5b: General Solution for ODEs =

Problem Statement
Problem found in the textbook on page 59, problem 5. Find a general solution. Check your answer by substitution.


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

$$\displaystyle {y}''+2\pi {y}'+\pi^{2}y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.1)
 * }

Solution
Using the following:


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$$\displaystyle {y}'=\frac{dy}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.2)
 * }


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

$$\displaystyle {y}''=\frac{d^{2}y}{dx^{2}}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.3)
 * }

The given equation can be rewritten as follows, using (1.5b.2) and (1.5b.3):
 * {| style="width:100%" border="0"

$$\displaystyle \frac{d^{2}y}{dx^{2}}+2\pi\frac{dy}{dx}+\pi^{2}y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.4)
 * }

Substituting
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$$\displaystyle D= \frac{d}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.5)
 * }

The ODE can be expressed as follows:
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$$\displaystyle (D^{2}+2\pi D+\pi^{2})y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.6)
 * }

Solving
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$$\displaystyle (D^{2}+2\pi D+\pi^{2})=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.7)
 * }


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

$$\displaystyle (D+\pi)^{2}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.8)
 * }


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

$$\displaystyle D=-\pi,-\pi$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.9)
 * }

The final general solution for this problem is:

Reach the general solution :


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


 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.10)
 * }

Check by substitution:

From (1.5b.10), we find,
 * {| style="width:100%" border="0"

$$\displaystyle {y}'=-\pi(C_{1}+C_{2}x)e^{-\pi x}+C_{2}e^{-\pi x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.11)
 * }


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

$$\displaystyle {y}''=-\pi(-\pi C_{1}+-\pi C_{2}x)e^{-\pi x}-\pi C_{2}e^{-\pi x}-\pi C_{2}e^{-\pi x}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.12)
 * }

Substituting (1.5b.11) and (1.5b.12) in (1.5b.1), we reach:
 * {| style="width:100%" border="0"

$$\displaystyle (\pi^{2}(C_{1}+C_{2}x)e^{-\pi x}-2\pi C_{2}e^{-\pi x})+2\pi(-\pi(C_{1}+C_{2}x)e^{-\pi x}+C_{2}e^{-\pi x})+\pi^{2}((C_{1}+C_{2}x)e^{-\pi x})=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.13)
 * }


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

$$\displaystyle \pi^{2}C_{1}e^{-\pi x}+\pi^{2}C_{2}xe^{-\pi x} -2\pi C_{2}e^{-\pi x}-2\pi^{2}C_{1}e^{-\pi x} -2\pi^{2} C_{2}xe^{-\pi x}+2\pi C_{2}e^{-\pi x} +\pi^{2}C_{1}e^{-\pi x} +\pi^{2}C_{2}xe^{-\pi x}=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.14)
 * }


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

$$\displaystyle 0=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.5b.15)
 * }

Therefore the solution is correct.

= Problem 6: Classification of ODEs and the Superposition Prcinciple =

Problem Statement
For each ODE in Fig.2 in Kreyszig 2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied.

Order
The order of an equation is determined by the highest derivative. In this report, the first derivative of the $$\displaystyle y$$ variable is denoted as $$\displaystyle y'$$, and the second derivative is denoted as $$\displaystyle y''$$, and so on. This can be determined upon observation of the equation.

Linearity
An ordinary differential equation (ODE) is considered linear if it can be brought to the form :


 * $$y'+p(x)y=q(x)\!$$

Superposition
Superposition can be applied if when a homogeneous and particular solution of an original equation are added, that they are equivalent to the original equation. In this report, variables with a bar over them represent the addition of the homogeneous and particular solution's same variable.


 * For example:


 * $$ y_p+y_h=\overline y \!$$

Given
The following equations were given in the textbook on p. 3 :
 * {| style="width:100%" border="0"

$$\displaystyle y''=g=constant$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6a.1)
 * }


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

$$\displaystyle mv'=mg-bv^2$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6b.1)
 * }


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

$$\displaystyle h'=-k\sqrt{h}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6c.1)
 * }


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

$$\displaystyle my''+ky=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6d.1)
 * }


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

$$\displaystyle y''+\omega_0^2 y=\cos\omega t, \omega_0 = \omega$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6e.1)
 * }


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

$$\displaystyle LI''+RI'+\frac{1}{C} I=E'$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6f.1)
 * }


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

$$\displaystyle EIy^{\omega}=f(x)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6g.1)
 * }


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

$$\displaystyle L\theta''+g\sin\theta=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6h.1)
 * }

1.6a

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

$$\displaystyle y''=g=constant$$ Order: 2nd Linear: yes Superposition: yes The given equation can be algebraically modified as the following:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6a.1)
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y''=g\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6a.1)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle y_{h}''=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6a.2)
 * }


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

$$\displaystyle y_{p}''=g\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6a.3)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (y_{h}+y_{p})=\overline{y}''\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6a.4)
 * }

The solution resembles the original equation, therefore superposition is possible

1.6b

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

$$\displaystyle mv'=mg-bv^2$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6b.1)
 * }

Order: 1st Linear: no Superposition: no The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle mv'+bv^2=mg\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6b.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle mv_{h}'+bv_{h}^2=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6b.3)
 * }


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

$$\displaystyle mv_{p}'+bv_{p}^2=mg\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6b.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle m(v_{h}'+v_{p}')+b(v_{h}^2+v_{p}^2)\not= m\overline{v}'+b(\overline{v}^2)\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6b.5)
 * }

The solution cannot be algebraically modified to resemble the original equation, and therefore superposition is NOT possible as also proven in the class notes

1.6c

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

$$\displaystyle h'=-k\sqrt{h}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6c.1)
 * }

Order: 1st Linear: no Superposition: no The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle h'+k\sqrt{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6c.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle h_{h}'+k\sqrt{h_{h}}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6c.3)
 * }


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

$$\displaystyle h_{p}'+k\sqrt{h_{p}}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6c.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (h_{h}'+h_{p}')+k(\sqrt{h_{h}}+\sqrt{h_{p}})\not=\overline{h}'+k\sqrt{\overline {h}}\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6c.5)
 * }

The solution cannot be algebraically modified to resemble the original equation, and therefore superposition is NOT possible.

1.6d

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

$$\displaystyle my''+ky=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6d.1)
 * }

Order: 2nd Linear: yes Superposition: yes The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle y''+\frac{k}{m}y=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6d.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle y_{h}''+\frac{k}{m}y_{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6d.3)
 * }


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

$$\displaystyle y_{p}''+\frac{k}{m}y_{p}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6d.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (y_{h}+y_{p})+\frac{k}{m}(y_{h}+y_{p})=\overline{y}''+\frac{k}{m}\overline{y}\!$$ The solution resembles the original equation, therefore superposition is possible
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6d.5)
 * }

1.6e

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

$$\displaystyle y''+\omega_0^2 y=\cos\omega t, \omega_0 = \omega$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6e.1)
 * }

Order: 2nd Linear: yes Superposition: yes

The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle y''+\omega _{o}^{2}y=\cos (\omega t)\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6e.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle y_{h}''+\omega _{o}^{2}y_{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6e.3)
 * }


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

$$\displaystyle y_{p}''+\omega _{o}^{2}y_{p}=\cos (\omega t)\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6e.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (y_{h}+y_{p})+\omega _{o}^{2}(y_{h}+y_{p})=\overline{y}''+\omega _{o}^{2}\overline{y}\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6e.5)
 * }

The solution resembles the original equation, therefore superposition is possible

1.6f

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

$$\displaystyle LI''+RI'+\frac{1}{C} I=E'$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6f.1)
 * }

Order: 2nd linear: yes superposition: yes

The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle I''+\frac{R}{L}I'+\frac{1}{LC}I=\frac{E'}{L}\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6f.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle I_{h}''+\frac{R}{L}I_{h}'+\frac{1}{LC}I_{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6f.3)
 * }


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

$$\displaystyle I_{p}''+\frac{R}{L}I_{p}'+\frac{1}{LC}I_{p}=\frac{E'}{L}\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6f.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (I_{h}+I_{p})+\frac{R}{L}(I_{h}'+I_{p}')+\frac{1}{LC}(I_{h}+I_{p})=\overline{I}''+\frac{R}{L}\overline{I}'+\frac{1}{LC}\overline{I}\!$$ The solution resembles the original equation, therefore superposition is possible
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6f.5)
 * }

1.6g

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

$$\displaystyle EIy=f(x)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6g.1)
 * }

Order: 4th Linear: yes Superposition: yes

The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle y-\frac{1}{EI}y=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6g.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle y_{h}-\frac{1}{EI}y_{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6g.3)
 * }


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

$$\displaystyle y_{p}-\frac{1}{EI}y_{p}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6g.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (y_{h}'+y_{p}')-\frac{1}{EI}(y_{h}+y_{p})=\overline{y}-\frac{1}{EI}\overline{y}\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6g.5)
 * }

The solution resembles the original equation, therefore superposition is possible

1.6h

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

$$\displaystyle L\theta''+g\sin\theta=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6h.1)
 * }

Order: 2nd Linear: no Superposition:no

The given equation can be algebraically modified as the following:
 * {| style="width:100%" border="0"

$$\displaystyle \theta''+\frac{g}{L}\sin\theta=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6h.2)
 * }

It can be split up into the following homogeneous and particular solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \theta_{h}''+\frac{g}{L}\sin\theta_{h}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6h.3)
 * }


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

$$\displaystyle \theta_{p}''+\frac{g}{L}\sin\theta_{p}=0\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6h.4)
 * }

Adding the two solutions:
 * {| style="width:100%" border="0"

$$\displaystyle (\theta_{h}+\theta_{p})+\frac{g}{L}(\sin\theta_{h}+\sin\theta_{p})\not=\overline{\theta}''+\frac{g}{L}\sin\overline{\theta}\!$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(1.6h.5)
 * }

The solution cannot be algebraically modified to resemble the original equation, and therefore superposition is NOT possible.

=Notes and References=