# Intermediate engineering analysis/Problem set 1

## R.1.1: Equation of Motion for a Spring-dashpot Parallel System in Series with Mass and Applied Force

### Problem Statement

Given a spring-dashpot system in parallel with an applied force, find the equation of motion.

### Background Theory

For this problem, Newton's second law is used,

$\sum F=ma$

(1-1)

and applied to the mass at the end of a spring and damper in parallel.

### Solution

Assuming no rotation of the mass,

$y=y_{m}=y_{k}=y_{c}$

(1-2)

Therefore, the spring and damper forces can be written as, respectively,

$f_{k}=ky$

(1-3)

and

$f_{c}=cy'$

(1-4)

And the resultant force on the mass can be written as

$F=ma=my''$

Now, from equations (1-3 ) and (1-4 ) all of the forces can be substituted into Newton's second law, and since each distance is equal,

$\sum F=my''=-f_{k}-f_{c}+f(t)$

(1-5)

Substituting these exact force equations,

$my''=-ky-cy'+f(t)$

A little algebraic manipulation yields

$my''+cy'+ky=f(t)$

Finally, dividing by the mass to put the equation in standard form gives the final equation of motion for the mass:

$y''+{\frac {c}{m}}y'+{\frac {k}{m}}y={\frac {1}{m}}f(t))$

(1-6)

## R.1.2: Equation of Motion for a Spring-mass-dashpot System with Applied Force

### Problem Statement

For this problem, the task was to derive the equation of motion of the spring-mass-dashpot in Fig. 53, in K 2011 p.85 with an applied force r(t).

### Background Theory

To solve this problem, note that the ODE of a damped mass-spring system is

$my''+cy'+ky=0$

(2-1)

When there is a external formal added to the model r(t) on the right. This then gives up the equation

$my''+cy'+ky=r(t)$

(2-2)

### Solution

In this problem referencing back to equation (1-2 )

$y=y_{k}=y_{c}$

The resultant force for the system can be described as stated in equation (1-5 )

$\sum F=my''=-f_{k}-f_{c}+f(t)$

Since r(t) is the external force,

$f(t)=r(t)$

The model of a mass-spring system ODE with external force on the right is modeled as

$my''+f_{internal}=r(t)$

The internal force is equal to the force of the spring making this equation

$my''+ky=r(t)$

When a dashpot is added, the force of the dashpot cy’ is added to the equation making it

$my''+cy'+ky=r(t)$

(2-3)

For the characteristic equation m is divided throughout the whole equation making it

(2-4)

## R.1.3: Free-body Diagram and Equation of Motion for Spring-dashpot-mass System

### Problem Statement

For this problem we were to draw the FBDs and derive the equation of motion for the following system.

### Background Theory

To start solving this problem, we have to first look at how the spring, dashpot, and mass interact with each other by analyzing their respective free-body diagrams.

Free-Body Diagram Derived Equation
$\sum F{_{y}}=m{\frac {dv}{dt}}\rightarrow ma=my''$

and
$my''=f(t)-f_{I}$

$ky'=f_{c}(t)-f_{k}(t)$

$cy=f_{I}-f_{c}(t)$

Next we take a look at what we already know.

We know from kinematics that:

$y=y{_{k}}+y{_{c}}$

(3-1)

We also know from kinetics that:

$my''+f{_{internal}}=f(t)$

(3-2)

$f{_{i}}=f{_{k}}=f{_{c}}$

(3-3)

From the constitutive relations we know:

$f_{k}=ky_{k}$

(3-4)

$f{_{c}}=cy{_{c}}'$

(3-5)

We also know:

$y''=y{_{k}}''+y{_{c}}''$

(3-6)

$f(k)=f(c)\Rightarrow ky{_{k}}=cy{_{c}}'$

(3-7)

by solving for $y_{c}'$  we get

$y_{c}'={\frac {k}{c}}y_{k}$

(3-8)

### Solution

To derive the equation of motion we have to manipulate and combine a few formulas that we know. We first take what we found in (3-2 )

and substitute in

$y{_{k}}''+y{_{c}}''$

(3-9)

for $y''$ . From equations (3-3 ), (3-4 ), and (3-5

) we can substitute $f{_{i}}$  for

$ky{_{k}}+cy{_{c}}'$

After the substitutions into equation (3-2 ) we have

$m(y{_{k}}''+y{_{c}}'')+ky{_{k}}+cy{_{c}}'=f(t)$

(3-10)

From the equation (3-8

) we know

$y_{c}'={\frac {k}{c}}y_{k}$

We can then substitute this into equation (3-10 ) getting

$m(y{_{k}}''+{\frac {k}{c}}y{_{k}}')+ky{_{k}}+ky{_{k}}=f(t)$

To get the answer in its final form we divide both sides of the equation by m, and our final answer is

$y{_{k}}''+{\frac {k}{c}}y{_{k}}'+2{\frac {k}{m}}y{_{k}}={\frac {1}{m}}f(t)$

(3-11)

## R.1.4: Derivation of Voltage-Charge and Voltage-Current Relationships

### Problem Overview

For this problem, the goal is to derive two different equations from the circuit equation

$V=LC{\frac {d^{2}v_{c}}{dt^{2}}}+RC{\frac {dv_{c}}{dt}}+v_{c}$

(4-1)

Each separate derivation is presented in Voltage-Charge Derivation and Voltage-Current Derivation

### Background Theory

To solve these derivation, it is wise to note that capacitance is

$Q=Cv_{c}$

(4-2)

It can also be noted that

$Q=\int idt$

This means that

$\int idt=Cv_{C}$

Completing the integration results in

$I=C{\frac {dv_{C}}{dt}}$

(4-3)

It should also be noted that (4-2 ) can be written in the form

$v_{c}={\frac {1}{C}}Q$

(4-4)

It should also be noted that (4-3 ) can be written in the form

${\frac {dv_{C}}{dt}}={\frac {1}{C}}I$

(4-5)

### Voltage-Charge Derivation

#### Problem Statement

For this problem, the Voltage-Charge equation

$V=LQ''+RQ'+{\frac {1}{C}}Q$

(4-6)

are derived from (4-1 )

#### Solution

Taking the derivative of (4-2 ) are taken with respect to time

$Q'=C{\frac {dv_{c}}{dt}}$

(4-7)

$Q''=C{\frac {d^{2}v_{c}}{dt^{2}}}$

(4-8)

Substituting (4-4 ), (4-7 ), and (4-8 ) into (4-1 ) results in (4-6 ) such that

$LC{\frac {d^{2}v_{c}}{dt^{2}}}+RC{\frac {dv_{c}}{dt}}+v_{c}=LQ''+RQ'+{\frac {1}{C}}Q$

(4-9)

### Voltage-Current Derivation

#### Problem Statement

For this problem, the following equation

$V'=LI''+RI'+{\frac {1}{C}}I$

(4-10)

are derived from (4-1 )

#### Solution

The first step is to take the derivative of (4-1 ) resulting in

$V'=LC{\frac {d^{3}v_{c}}{dt^{3}}}+RC{\frac {d^{2}v_{c}}{dt^{2}}}+{\frac {dv_{c}}{dt}}$

(4-11)

Taking the derivative of (4-3 ) are taken with respect to time

$I'=C{\frac {d^{2}v_{c}}{dt^{2}}}$

(4-12)

$I''=C{\frac {d^{3}v_{c}}{dt^{3}}}$

(4-13)

Substituting (4-3 ), (4-12 ), and (4-13 ) into (4-11 ) results in (4-10 ) such that

$LC{\frac {d^{3}v_{c}}{dt^{3}}}+RC{\frac {d^{2}v_{c}}{dt^{2}}}+{\frac {dv_{c}}{dt}}=LI''+RI'+{\frac {1}{C}}I$

(4-16)

## R.1.5: Solutions of General 2nd Order ODEs

### Background Theory

Consider a second-order homogeneous linear ODE with constant coefficients a and b.

$y''+ay'+by=0$

(5-1)

To solve this problem, note that the solution to a first-order linear ODE of the form:

$y'+ky=0$

is an exponential function, yielding a solution of the form:

$y=e^{\lambda x}$

(5-2)

Based upon this, we will expand the solution to apply to second-linear homogeneous ODEs. To do so, we must first find equations describing $y'$  and $y''$ , and so we will take the derivatives (with respect to x) of (5-2 ).

$y'=\lambda e^{\lambda x}$

(5-3)

$y''=\lambda ^{2}e^{\lambda x}$

(5-4)

We will now substitute (5-2 ), (5-3 ), and (5-4 ) into (5-1 ) to obtain the relationship:

$\lambda ^{2}e^{\lambda x}+a\lambda e^{\lambda x}+be^{\lambda x}=0$

Simplifying, we have:

$(\lambda ^{2}+a\lambda +b)e^{\lambda x}=0$

$\lambda ^{2}+a\lambda +b=0$

(5-5)

Since (5-5 ) follows the same form as the quadratic equation, we can solve for $\lambda _{1}$  and $\lambda _{2}$  as follows:

$\lambda _{1}={\frac {1}{2}}(-a+{\sqrt {(a^{2}-4b)}}$

$\lambda _{2}={\frac {1}{2}}(-a-{\sqrt {(a^{2}-4b)}}$

Referring back to algebra, we know that the solution to these two equations can be one of three cases:

#### Case 1

Two real roots if...

$a^{2}-4b>0$

These two roots give us two solutions:

 $y_{1}=e^{\lambda _{1}x}$ and $y_{2}=e^{\lambda _{2}x}$ The corresponding general solution then takes the form of the following:

$y=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}$

(5-6)

#### Case 2

A real double root if...

$a^{2}-4b=0$

This yields only one solution:

$y_{1}=e^{\lambda x}$

In order to form a basis, a set of two linearly independent solutions, we must find another solution. To do so, we will use the method of reduction of order, where:

$y_{2}=uy_{1}$
$y'_{2}=u'y_{1}+uy'_{1}$
and
$y''_{2}=u''y_{1}+2u'y'_{1}+uy''_{1}$
yeilding
$(u''y_{1}+2u'y'_{1}+uy''_{1})+a(u'y_{1}+uy'_{1})+buy_{1}=0$
$u''y_{1}+u'(2y'_{1}+ay_{1})+u(y''_{1}+a'y_{1}+by_{1})=0$

Since $y_{1}$  is a solution of (5-1), the last set of parentheses is zero.
Similarly, the first parentheses is zero as well, because

$2y'_{1}=-ae^{ax/2}=-ay_{1}$

From integration, we get the solution:

$u=c_{1}x+c_{2}$

If we set $c_{1}=1$  and $c_{2}=0$

$y_{2}=xy_{1}$

Thus, the general solution is:

$y=(c_{1}x+c_{2})e^{\lambda _{2}x}$

(5-7)

#### Case 3

Complex conjugate roots if...

$a^{2}-4b<0$

In this case, the roots of (5-5 ) are complex:

$y_{1}=e^{-ax/2}cos(\omega x)$
and
$y_{2}=e^{-ax/2}sin(\omega x)$

Thus, the corresponding general solution is of the form:

$y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))$

(5-8)

### P.2.2.4

#### Problem Statement

For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:

$y''+4y'+(\pi ^{2}+4)y=0$

(5a-1)

#### Solution

Following the process that yields (5-5 ), we find the equation:

$\lambda ^{2}e^{\lambda x}+4\lambda e^{\lambda x}+(\pi ^{2}+4)e^{\lambda x}=0$

(5a-2)

To find the form of the general solution, we must solve for the values of $\lambda$  using the quadratic formula:

File:224 solution.png

Since the solutions to $\lambda$  are complex numbers, our solution takes the form of (5-8 ):

$y=e^{-2x}(Acos(\pi x)+Bsin(\pi x)),$

(5a-3)

A and B are constant coefficients of unknown value. Had we been given initial values of $y(x)$  and $y'(x)$ , we would solve for those values as well.

#### Confirmation of Solution

To confirm that the solution is true, we will substitute $y$ , $y'$ , and $y''$  into (5a-1 ). If the final result is a tautology then the solution is confirmed. If the result is a contradiction, a statement that is never true, then our solution is disproved.

$y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))$

$y'=-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))$

(5a-4)

$y''=4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-2e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(-A\pi sin(\pi x)$

$+B\pi cos(\pi x))+e^{-2x}(-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x))$

(5a-5)

Substituting (5a-3 ), (5a-4 ), and (5a-5 ) into (5a-1 ), we are left with:

$[4e^{-2x}(Acos(\pi x)+Bsin(\pi x))-2e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))+e^{-2x}(-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x))]$

$+4[-2e^{-2x}(Acos(\pi x)+Bsin(\pi x))+e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))]+(\pi ^{2}+4)[e^{-2x}(Acos(\pi x)+Bsin(\pi x))]=0$

Combining similar terms allows us to clean up this solution and check our answer:

$e^{-2x}[cos(\pi x)[(4A-2B\pi -2B\pi -A\pi ^{2})+(-8A+4B\pi )+(A\pi ^{2}+4A)]+sin(\pi x)[(4B+2A\pi +2A\pi -B\pi ^{2})+(-8B-4A\pi )+(B\pi ^{2}+4B)]]=0$

$e^{-2x}[cos(\pi x)[A(4+4-8-\pi ^{2}+\pi ^{2})+B(-2\pi -2\pi +4\pi )]+sin(\pi x)[A(2\pi +2\pi -4\pi )+B(4+4-8-\pi ^{2}+\pi ^{2})]]=0$

$e^{-2x}[cos(\pi x)(0)+sin(\pi x)(0)]=0$

$0=0$

#### Final Solution

Since the outcome of our check is a tautology, the solution is confirmed for all values of A and B and we affirm that the final solution is:

$y=e^{-2x}(Acos(\pi x)+Bsin(\pi x))$

### P.2.2.5

#### Problem Statement

For this problem, we were tasked with finding the general solution to the homogeneous linear ODE given below:

${y}''+2\pi {y}'+\pi ^{2}y=0$

(5b-1)

#### Solution

Following the process that yields (5-5 ), we find the equation:

$\lambda ^{2}e^{\lambda x}+2\pi \lambda e^{\lambda x}+\pi ^{2}e^{\lambda x}=0$

(5b-2)

To find the form of the general solution, we must solve for the values of $\lambda$ . Instead of going through the quadratic formula, we can analyze the discriminant.

${\sqrt {(2\pi )^{2}-4\pi ^{2}}}=0$

Thus we see that the discriminant follows Case 2 and the general solution must follow the form of (5-7 ), and we have the following:

$y=(c_{1}+c_{2}x)e^{-\pi x}$

(5b-3)

where $c_{1}$  and $c_{2}$  are unknown constant coefficients. If this were an initial value problem with values of $y(x)$  and $y'(x)$ , we would solve for those values as well.

#### Confirmation of Solution

To confirm that the solution is true, we will substitute $y$ , $y'$ , and $y''$  into (5a-1 ) and check to see if the final result is a tautology or a contradiction.

$y'=-\pi (c_{1}+c_{2}x)e^{-\pi x}+c_{2}e^{-\pi x}$

(5b-4)

$y''=\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+-2\pi c_{2}e^{-\pi x}$

(5b-5)

Substituting (5b-3 ), (5b-4 ), and (5b-5 ) into (5b-1 ), we are left with:

$[\pi ^{2}(c_{1}+c_{2}x)e^{-\pi x}+-2\pi c_{2}e^{-\pi x}]+2\pi [-\pi (c_{1}+c_{2}x)e^{-\pi x}+c_{2}e^{-\pi x}]+\pi ^{2}[(c_{1}+c_{2}x)e^{-\pi x}]=0$

After combining similar terms, we have the form:

$\pi ^{2}[(c_{1}+c_{2}x)e^{-\pi x}-2(c_{1}+c_{2}x)e^{-\pi x}+(c_{1}+c_{2}x)e^{-\pi x}]+2\pi [c_{2}e^{-\pi x}-c_{2}e^{-\pi x}]=0$

$0=0$

#### Final Solution

Since the above result is a tautology for all values of $c_{1}$  and $c_{2}$ , (5-3 ) is confirmed and the general solution to (5-1 ) is:

$y=(c_{1}+c_{2}x)e^{-\pi x}$

### References

1. EGM4313: Section 2 Notes (.djvu)
2. Kreyszig, Erwin (2011). Advanced Engineering Mathematics. John Wily & Sons, Inc. ISBN 978-0-470-45836-5. Unknown parameter |coauthors= ignored (|author= suggested) (help) Pg 52-58
3. Introduction to Propositional Logic

## R.1.6: Determination of ODE Order, Linearity, and Application of the Superposition Principle

### Problem Statement

Given the eight Ordinary Differential Equations shown below:
(1) Determine the order
(2) Determine the linearity (or lack of)
(3) Show whether the principle of superposition can be applied for each equation

### Background Theory

Consider a function ${\bar {y}}(x)$ , defined as the sum of the homogeneous solution, $y_{h}(x)$ , and the particular solution, $y_{p}(x)$ :

${\bar {y}}(x)=y_{p}(x)+y_{h}(x)$

(6-1)

Example:

Consider the Differential Equation in standard form:

$y''+p(x)y'+q(x)y=r(x)$

(6-2)

The homogeneous solution is then:

$y_{h}''+p(x)y_{h}'+q(x)y_{h}=0$

(6-3)

and the particular solution is:

$y_{p}''+p(x)y_{p}'+q(x)y_{p}=r(x)$

(6-4)

Now we sum up equations (6-3 ) and (6-4 ) to get:

$(y_{h}''+y_{p}'')+p(x)(y_{h}'+y_{p}')+q(x)(y_{h}+y_{p})=r(x)$

(6-5)

If the Equation is Linear then some reductions can take place:

$y_{h}''+y_{p}''=(y_{h}+y_{p})''={\bar {y}}''$

and

$y_{h}'+y_{p}'=(y_{h}+y_{p})'={\bar {y}}'$

Now we can make some substitutions into (6-5 )using (6-3 )and (6-4 ):

${\bar {y}}''+p(x){\bar {y}}'+q(x){\bar {y}}=r(x)$

(6-6)

Which is the same as equation (6-2 ):

$y''+p(x)y'+q(x)y=r(x)$

### Solution

#### Part A

$y''=g=constant$

(6a-1)

Order: $2^{nd}$  Order ODE
Linearity: Linear
Superposition: Yes

This can be shown using the method from the background theory.

Recall:

${\bar {y}}(x)=y_{p}(x)+y_{h}(x)$

The homogeneous solution is:

$y_{h}''=0$

(6a-2)

The particular solution is:

$y_{p}''=g$

(6a-3)

Now we add Equations (6a-2 ) and (6a-3 ):

$y_{h}''+y_{p}''=g$

(6a-4)

We can use (6-1 ) to simplify (6a-4 ):

$(y_{h}+y_{p})''=g$

(6a-5)

${\bar {y}}''=g$

(6a-6)

Since (6a-1 ) and (6a-6 ) are the same then we can apply Superposition.

#### Part B

$mv'=mg-bv^{2}$

(6b-1)

Order: $1^{st}$  Order ODE
Linearity: Non-Linear
Superposition: No

This can be shown using the method from the background theory.
Before we start it is easier to do the problem if the equation is rearranged as such:

$mv'+bv^{2}=mg$

(6b-2)

Recall:

${\bar {v}}(x)=v_{p}(x)+v_{h}(x)$

The homogeneous solution is:

$mv_{h}'+mv_{h}^{2}=0$

(6b-3)

The particular solution is:

$mv_{p}'+mv_{p}^{2}=mg$

(6b-4)

Now we add Equations (6b-3 ) and (6b-4 ):

$mv_{h}'+mv_{h}^{2}+mv_{p}'+mv_{p}^{2}=mg$

(6b-5)

We can use (6-1 ) to simplify (6b-5 ):

$m(y_{h}+y_{p})'+m(v_{h}^{2}+v_{p}^{2})=mg$

(6b-6)

Note that:

$(v_{h}^{2}+v_{p}^{2})\neq {\bar {v}}^{2}$

So we cannot apply Super position.

#### Part C

${h}'=-k{\sqrt {h}}$

(6c-1)

Order: $1^{st}$  Order ODE
Linearity: Non-Linear
Superposition: No

This can be shown using the method from the background theory.
Before we start it is easier to do the problem if the equation is rearranged like this:

${h}'+k{\sqrt {h}}=0$

(6c-2)

Recall:

${\bar {h}}(x)=h_{h}(x)+h_{p}(x)$

The homogeneous solution is:

${h}_{h}'+k{\sqrt {h_{h}}}=0$

(6c-3)

The particular solution is:

${h_{p}}'+k{\sqrt {h_{p}}}=0$

(6c-4)

Now we add Equations (6c-3 ) and (6c-4 ):

${h_{h}}'+k{\sqrt {h_{h}}}+{h_{p}}'+k{\sqrt {h_{p}}}=0$

(6c-5)

We can use (6-1 ) to simplify (6c-5 ):

$({h_{h}}+{h_{p}})'+k({\sqrt {h_{h}}}+{\sqrt {h_{p}}})=0$

(6c-6)

Note that:

$({\sqrt {h_{h}}}+{\sqrt {h_{p}}})\neq {\sqrt {\bar {h}}}$

So we cannot apply Super position.

#### Part D

$m{y}''+ky=0$

(6d-1)

Order: $2^{nd}$  Order ODE
Linearity: Linear
Superposition: Yes

This can be shown using the method from the background theory.

Recall:

${\bar {y}}(x)=y_{p}(x)+y_{h}(x)$

The homogeneous solution is:

$m{y}_{h}''+ky_{h}=0$

(6d-2)

The particular solution is:

$m{y}_{p}''+ky_{p}=0$

(6d-3)

Now we add Equations (6d-2 ) and (6d-3 ):

$m{y}_{h}''+ky_{h}+m{y}_{p}''+ky_{p}=0$

(6d-4)

We can use (6-1 ) to simplify (6d-4 ):

$m({y}_{h}+{y}_{p})''+k(y_{h}+y_{p})=0$

(6d-5)

$m{\bar {y}}''+k{\bar {y}}=0$

(6d-6)

Since (6d-1 ) and (6d-6 ) are the same then we can apply Superposition.

#### Part E

${y}''+\omega _{0}^{2}y=\cos \omega t,\omega _{0}\approx \omega$

(6e-1)

Order: $2^{nd}$  Order ODE
Linearity: Linear
Superposition: Yes

This can be shown using the method from the background theory.

Recall:

${\bar {y}}(x)=y_{p}(x)+y_{h}(x)$

The homogeneous solution is:

${y}_{h}''+\omega _{0}^{2}y_{h}=0$

(6e-2)

The particular solution is:

${y}_{p}''+\omega _{0}^{2}y_{p}=\cos \omega t$

(6e-3)

Now we add Equations (6e-2 ) and (6e-3 ):

${y}_{h}''+\omega _{0}^{2}y_{h}+{y}_{p}''+\omega _{0}^{2}y_{p}=\cos \omega t$

(6e-4)

We can use (6-1 ) to simplify (6e-4 ):

$({y}_{h}+{y}_{p})''+\omega _{0}^{2}(y_{h}+y_{p})=\cos \omega t$

(6e-5)

${\bar {y}}''+\omega _{0}^{2}{\bar {y}}=\cos \omega t$

(6e-6)

Since (6e-1 ) and (6e-6 ) are the same then we can apply Superposition.

#### Part F

$LI''+RI'+{\frac {1}{C}}I=E'$

(6f-1)

Order: $2^{nd}$  Order ODE
Linearity: Linear
Superposition: Yes

This can be shown using the method from the background theory.

Recall:

${\bar {I}}(x)=I_{p}(x)+I_{h}(x)$

The homogeneous solution is:

$LI_{h}''+RI_{h}'+{\frac {1}{C}}I_{h}=0$

(6f-2)

The particular solution is:

$LI_{p}''+RI_{p}'+{\frac {1}{C}}I_{p}=E'$

(6f-3)

Now we add Equations (6f-2 ) and (6f-3 ):

$LI_{h}''+RI_{h}'+{\frac {1}{C}}I_{h}+LI_{p}''+RI_{p}'+{\frac {1}{C}}I_{p}=E'$

(6f-4)

We can use (6-1 ) to simplify (6f-4 ):

$L(I_{h}+I_{p})''+R(I_{h}+I_{p})'+{\frac {1}{C}}(I_{h}+I_{p})=E'$

(6f-5)

$L{\bar {I}}''+R{\bar {I}}'+{\frac {1}{C}}{\bar {I}}=E'$

(6f-6)

Since (6f-1 ) and (6f-6 ) are the same then we can apply Superposition.

#### Part G

$EIy^{iv}=f(x)$

(6g-1)

Order: $4^{th}$  Order ODE
Linearity: Linear
Superposition: Yes

This can be shown using the method from the background theory.

Recall:

${\bar {y}}(x)=y_{p}(x)+y_{h}(x)$

The homogeneous solution is:

$EIy_{h}^{iv}=0$

(6g-2)

The particular solution is:

$EIy_{p}^{iv}=f(x)$

(6g-3)

Now we add Equations (6g-2 ) and (6g-3 ):

$EIy_{h}^{iv}+EIy_{p}^{iv}=f(x)$

(6g-4)

We can use (6-1 ) to simplify (6g-4 ):

$EI(y_{h}+y_{p})^{iv}=f(x)$

(6g-5)

$EI({\bar {y}})^{iv}=f(x)$

(6g-6)

Since (6g-1 ) and (6g-6 ) are the same then we can apply Superposition.

#### Part H

$L\theta ''+g\sin \theta =0$

(6h-1)

Order: $2^{nd}$  Order ODE
Linearity: Non-Linear
Superposition: No

This can be shown using the method from the background theory.

Recall:

${\bar {\theta }}(x)=\theta _{p}(x)+\theta _{h}(x)$

The homogeneous solution is:

$L\theta _{h}''+g\sin \theta _{h}=0$

(6h-2)

The particular solution is:

$L\theta _{p}''+g\sin \theta _{p}=0$

(6h-2)

Now we add Equations (6h-2 ) and (6h-3 ):

$L\theta _{h}''+g\sin \theta _{h}+L\theta _{p}''+g\sin \theta _{p}=0$

(6h-4)

We can use (6-1 ) to simplify (6h-4 ):

$L(\theta _{h}+\theta _{p})''+g(\sin \theta _{h}+\sin \theta _{p})=0$

(6h-5)

Note that:

$(\sin \theta _{h}+\sin \theta _{p})\neq \sin {\bar {\theta }}$

So we cannot apply Super position.

### References

1. EGM4313: Section 2 Notes (.djvu)
2. Kreyszig, Erwin (2011). Advanced Engineering Mathematics. John Wily & Sons, Inc. ISBN 978-0-470-45836-5. Unknown parameter |coauthors= ignored (|author= suggested) (help) Pg 46-48