Mathematics AA HL

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I believe that learning takes place when there is a unison between theory and its application and IB has provided me the best opportunity to understand the practical application of mathematics. Since childhood I have been taught everything from a mathematical perspective and being an inquirer, I strongly believe the knowledge cannot be restricted by the barriers of curriculum.

Intrigued by differential equations which I had learnt in our curriculum, I decided to gather more information on this subject. Surprisingly, while talking to an acquaintance in my apartment regarding issues pertaining to the Control Systems, I discovered that a process which hasn’t been taught in our curriculum is used to solve the differential equations which are involved in the Control Systems in Engineering.

Thus, I wanted to gain more knowledge about this topic and had started researching on finding solutions to the differential equations that aren’t included in the curriculum. I have understood three new processes, such as Separation of Variable, Laplace's Transform and Runge Kutta Method, by attending a few courses and watching tutorial videos. I also discovered that these methods are only a continuation of the curriculum of the differential equation that is included in IB.

Hence, I have decided to work on this topic to evaluate the ways of finding roots of differential equations with selected practicality in the domain of Control System along with two separate methods and have discerned the surer way of calculation.

The prime objective of this exploration is to solve the differential equation procured from the transfer function (or output response) of the Rice Cooker during its testing phase using Integrating Factor Method and Laplace’s Method.

A rice cooker is a kitchen appliance which automatically cooks rice and is used widely by many households. It saves time and makes work easier. It also saves fuel (LPG) as it is powered by electricity.

A rice cooker uses a ratio of 2:1 for water and rice respectively in majority of the cases. It also has a thermometer which measures the temperature of the water as well as that of the rice inside. At 212℉ , the water turns to steam. The rice cooker switches to the warm mode or shuts off automatically depending on whether the rice has absorbed the water or not.

The rice cooker is an assembly of several components. Each element has its time complexity and limitation. Hence, each element's efficiency plays a significant role in the time complexity of the entire device. Transfer function is expressed mathematically, as follows:

Transfer Function = \(\frac{Laplace\ Transform\ of\ Output\ Signal}{Laplace\ Transform\ of\ Input\ Signal}\)

A differential equation is an equation with the derivatives of different orders of a function. A function is produced by the solution of a differential equation. This can further be used to anticipate the behavior of the original system . The degree of the derivative is called the order of the derivative.

Few examples are mentioned below:

\(a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0\)………(equation - 1)

Equation (1) is a second order differential equation

a , b, c are the coefficients of the second order derivative, first order derivative and the function itself respectively.

\(a\frac{dy}{dx}+by=0\)………(equation - 2)

Equation (2) is a first order differential equation

a, and b are the coefficients of the first order derivative and the function itself respectively.

In both the equations (1) and equation (2):

*y = function of *(*x*)

x = independent variable

Let us consider a linear differential equation of first order:

\(\frac{dy}{dx}\) + G(x) y = H(x)………(equation - 3)

G(x), and H(x) are the coefficients of the function (y) and the constant term respectively.

Integrating Factor (\(\gamma\)) of first order differential equation is given below:

\(γ=e^{\displaystyle\int G(x)dx}\)………(equation - 4)

The general solution of first order differential equation solved using Integrating Factor method is given below:

\(y × γ = \displaystyle\int N (x) × γdx\)………(equation - 5)

The integral transformation from ‘time domain’ (t) of a function to ‘complex frequency domain’ (s) of the function is known as Laplace Transform.

Formula:

L {f(t)} = F(s) = \(\displaystyle\int\limits^∞_0f(t)e^{-st}dt\) ………(equation - 6)

Where,

s = a + ib

a = Real value of s

b = Imaginary value of s

Of many methods, Method of Partial Fraction is generally used for performing Inverse Laplace Transform.

Following are the general solutions of ‘Inverse Laplace Transform:

\(L^{-1}\big\{\frac{p}{s+a}\big\}=be^{-at}\)

\(L^{-1}\bigg\{\frac{p}{(s+a)^n}\bigg\}=p\times\frac{t^{n-1}}{(n-1)!}\times e^{-at}\)

\(L^{-1}\big\{\frac{p}{s}\big\}=pt^{2}\)

Where,

a,p,n = constant

t = variable in time domain

s = variable in frequency domain

If an integral is in the form as shown below:

\(\displaystyle\int uvdx\)

*u, v = f *(*x*)

Then the solution of the integral by using formula of Integration by parts can be given by:

\(\displaystyle\int uvdx=u\displaystyle\int vdx-\bigg \{\displaystyle\int\frac{du}{dx}\bigg(\displaystyle\int vdx\bigg)dx\bigg\}\)

Data pertaining to the output response testing of rice cookers by commercial producers, is classified and not available to the public. This I found after searching considerably on the websites of reputed brands like Panasonic, Philips and Prestige. Thus, I have considered probable, hypothetical values to perform the calculations which can justify the actual process taking place in reality.

Considering the initial condition of the function to be zero, the ‘output response’ of the function is as follows:

\(\frac{dx}{dt}\) + 25x = 7t………(equation-7)

\(⟹\frac{dx}{dt}\) + 25x - 7t = 0

*x = ft =* output response of rice cooker with time (*t*)

**Solution by Method of Integrating Factor:**

Comparing equation (7) with equation (3):

*M *(*t *)*= *25

\(γ=e^{\ \int M(t)dt}\)

\(=e^{\int25\ dt}\)

= e^{25t}

From equation (5):

\(yγ=\displaystyle\int N(x)γdx\)

\(=>y×e^{25t}=\displaystyle\int7t×e^{25t}×dt\)

By the method of integration by parts, we consider that in the given integration:

\(\displaystyle\int 7t×e^{25t}×dt=\int uvdt\)

when,

*u = 7t*

\(v=e^{25t}\)

\(∴\displaystyle\int vdt=\int e^{25t}dt=\frac{e^{25t}}{25}\)

\(∴\frac{du}{dt}=\frac{d(7t)}{dt}=7\)

Using the formula of ‘Integration by Parts’:

\(\displaystyle\int7t×e^{25t}×dt\)

\(=7t×\displaystyle\int e^{25t}dt-\int\frac{d(7t)}{dt}×(\int e^{25t}dt)dt\)

\(=7t×\frac{e^{25t}}{25}+c_1-\displaystyle\int7×\frac{e^{25t}}{25}dt\)

\(=7t×\frac{e^{25t}}{25}+c_1-7×\displaystyle\int\frac{e^{25t}}{25}dt\)

\(=7t×\frac{e^{25t}}{25}+c_1-7×\displaystyle\int\frac{e^{12t}}{625}+c_2\)

where,

*c*_{1}*, c*_{2 }= integration constant

c_{1} + c_{2} = c = constant

\(∴\displaystyle\int7t×e^{25t}×dt=7t×\frac{e^{25t}}{25}-7×\frac{e^{25t}}{625}+c\)

\(=\frac{7e^{25t}}{25}\bigg[t-\frac{1}{25}\bigg]+c\)

Plugging the value of this integral in the solution by using the method of Integrating Factor:

\(y×e^{25t}=\frac{7e^{25t}}{25}\bigg[t-\frac{1}{25}\bigg]+c\)

The value of *c *(combined integration constant) of the above expression is taken to be zero for graphical representation.

*c = *0

By putting the value of *c* in the above equation:

\(y×e^{25t}=\frac{7e^{25t}}{25}\bigg[t-\frac{1}{25}\bigg]\)

\(y=\frac{7}{25}\bigg[t-\frac{1}{25}\bigg]\)………(equation - 8)

In Figure 1, The method of integrating factor has been used to find the solution of the differential equation.

The representation of the output response of the rice cooker with respect to time according to the equation can be seen in this graph. The shift in the output response of the rice cooker compared to time, which has been measured in seconds, is seen in graph one. Thus, time or independent variable is plotted along X – Axis and output response of the rice cooker or dependent variable is plotted along Y – Axis.

An increasing linear trend has been witnessed. As the operation time ranges from 0.00 seconds to 1.00 seconds, the output response increases from -0.0112 to 0.2688.

The obtained equation of trendline is shown below:

y = 0.28x - 0.0112

Where, y represents the output response of the Rice Cooker and x represents the time measured in seconds.

To find the output response of the device at initial condition, i.e., when (*t = *0), is shown below:

y = 0.28 × 0 - 0.0112

y = - 0.0112

Therefore, the output response of the device (rice cooker) before it is even started is equal to -0.0112.

This signifies that the

Moreover, to find the time at which the output response will be zero is shown below:

0 = 0.28x - 0.0112

\(⟹x=\frac{0.0112}{0.28}\)

⟹ x = 0.04 s

Thus, Output Response becomes 0 at time 0.04 s.

There are no significant points which are lying on either side of trendline. By observation, majority of the data points are found to lie on the trendline which strengthens the correlation as the number of outliers becomes zero. Furthermore, the strength of correlation is explained by a high value of regression coefficient (=1).

**Calculation of Processing Speed**

\(y=\frac{7}{25}\bigg[t-\frac{1}{25}\bigg]\)

Taking derivatives in both the sides with respect to ‘time’:

\(\frac{dy}{dt}=\frac{d}{dt}\bigg(\frac{7}{25}\bigg[t-\frac{1}{25}\bigg]\bigg)\)

\(=>\frac{dy}{dt}=\frac{7}{25}\bigg[\frac{d}{dt}(t)-\frac{d}{dt}\bigg(\frac{1}{25}\bigg)\bigg]\)

\(=>\frac{dy}{dt}=\frac{7}{25}[1-0]\)

\(=>\frac{dy}{dt}=\frac{7}{25}\)………(equation - 9)

In Figure 2, The method of integrating factor has been used to plot the solution of the differential equation. The representation of the output response of the rice cooker with respect to time according to the equation can be seen in this graph. Unlike Figure 1, Figure 2 represents the change in the processing speed of the rice cooker compared to its operation time which has been measured in seconds

Thus, time or independent variable is plotted along X – Axis and processing speed of the rice cooker or dependent variable is plotted along Y – Axis. A linear and constant trendline has been observed. As the operation time increases from 0.00 seconds to 1.00 seconds, the processing speed remains constant and equal to 0.28 s^{-1}.

The obtained equation of trendline is shown below:

y = 0.28

Here, *y *represents the processing speed (measured in s^{-1}) of the Rice Cooker and x represents the time in seconds.

**Solution by Laplace’s Method:**

Firstly, the Laplace Transform of both the sides of the equation (7) is obtained:

\(L\bigg\{\frac{dx}{dt}+25y\bigg\}=L\{7t\}\)

Applying the formula of Laplace Transform as per equation 6:

\(\displaystyle\int\limits^∞_0\bigg\{\frac{dx}{dt}+25x\bigg\}e^{-st}dt=\displaystyle\int\limits^∞_0\{7t\}e^{-st}dt\)

\(⟹\displaystyle\int\limits^∞_0\frac{dx}{dt}e^{-st}dt+25xe^{-st}dt=\displaystyle\int\limits^∞_0\{7t\}e^{-st}dt\)

\(⟹L\big\{\frac{dx}{dt}\big\}+25L\{x\}=L\{7t\}\)

\(⟹L\big\{\frac{dx}{dt}\big\}+25L\{x\}-L\{7t\}=0\) ………(equation - 10)

Let us consider each of the terms of the above equation (10) as individual terms as shown below:

\(Term\ 1=L\big\{\frac{dx}{dt}\big\}\)

\(Term\ 2=25L\{x\}\)

\(Term\ 3=L\{7t\}\)

Term 1:

\(L\{f(t)\}=\displaystyle\int\limits^∞_0f(t)e^{-st}dt=F(s)(say)\)

\(⟹L\bigg\{\frac{dx}{dt}\bigg\}=\displaystyle\int\limits^∞_0\frac{dx}{dt}e^{-st}dt\)

\(Let\ k=e^{-st}\)

\(\frac{dk}{dt}=\frac{d(e^{-st})}{dt}\)

= - se^{-st}

\(∴dk=-se^{-st}dt\)

Let *l* = x

\(\frac{dl}{dt}=\frac{d(x)}{dt}\)

\(⟹dl=\frac{dx}{dt}dt\)

Using the formula of ‘Integration by Parts’:

\(\displaystyle\int\limits^b_ak.l\ dt=k\displaystyle\int\limits^b_al\ dt-\displaystyle\int\limits^b_a\frac{dk}{dt}(\displaystyle\int\limits^b_al\ dt)dt\)

Now:

\(\displaystyle\int\limits^∞_0\frac{dx}{dt}e^{-st}dt=e^{-st}\displaystyle\int\limits^∞_0x\ dt-\displaystyle\int\limits^∞_0\frac{d(e^{-st})}{dt}(\displaystyle\int\limits^∞_0x\ dt)dt\)

\(=e^{-st}[f(∞)-f(0)]-\displaystyle\int\limits^∞_0-se^{-st}(\displaystyle\int\limits^∞_0x\ dt)dt\)

\(=e^{-st}[f(∞)-f(0)]+s\displaystyle\int\limits^∞_0e^{-st}(\displaystyle\int\limits^∞_0f\ (t))dt\)

= 0 + *sF*(*s*) = *sF*(*s*)

Considering the initial parameters to be zero [given,]

*Thus*, [*f*(**∞) **- *f*(0)] = 0

Term 2:

\(25\{x\}=25×\displaystyle\int\limits^∞_0xe^{-st}dt\)

\(=25×\displaystyle\int\limits^∞_0f(t)e^{-st}dt\)

= 25*F*(*s*)

Term 3:

\(L\{7t\}=\displaystyle\int\limits^∞_0\{7t\}e^{-st}dt=pqdt\)

where,

*p* = 7*t*

q = e^{-st}

\(\displaystyle\int\limits^∞_0pqdt\)

\(=p\displaystyle\int\limits^∞_0qdt-\displaystyle\int\limits^∞_0\frac{dp}{dt}×\bigg(\displaystyle\int\limits^∞_0qdt\bigg)dt\)

\(=7t\displaystyle\int\limits^∞_0e^{-st}dt-\displaystyle\int\limits^∞_0\frac{d(7t)}{dt}×\bigg(\displaystyle\int\limits^∞_0e^{-st}dt\bigg)dt\)

\(=7t×\frac{e^{-st}}{-s}\bigg|^∞_0-\displaystyle\int\limits^∞_07×\bigg(\frac{e^{-st}}{-s}\bigg|^∞_0\bigg)dt\)

\(=7t×\frac{e^{-st}}{-s}\bigg|^∞_0-7\frac{e^{-st}}{s^2}\bigg|^∞_0\)

\(=7×\bigg[t× \frac{e^{-st}}{-s}-\frac{e^{-st}}{s^2}\bigg|^∞_0\bigg]\)

\(=7×t× \frac{e^{-st}}{-s}-\frac{e^{-st}}{s^2}\bigg|^a_0\)

\(=7×\bigg[\frac{-ae^{-sa}}{s}-\frac{e^{-s×0}}{s^2}\bigg]-\bigg[0-\frac{1}{s^2}\bigg]\)

\(=7×\frac{1}{s^2}\)

Putting the values of Laplace Transform of all the terms in equation 10:

\(L\bigg\{\frac{dx}{dt}\bigg\}+25L\{y\}-L\{7t\}=0\)

\(=>sF(s)+25F(s)-7×\frac{1}{s^2}=0\)

\(=>sF(s)+25F(s)=\frac{7}{s^2}\)

\(=>(s+25)F(s)=\frac{7}{s^2}\)

\(=>F(s)=\frac{7}{s^2(s+25)}\)

To evaluate the Inverse Laplace Transform of this equation of output response of the Rice Cooker in complex frequency domain, the Method of Partial Fraction is used:

Assuming:

\(\frac{A}{s}+\frac{B}{s^2}+\frac{F}{s+25}=\frac{7}{s^2(s+25)}\) ………(equation 11)

A, B,C are constant

And, from the above equation, it can be told that:

(s), (s^{2}), and (s + 25) are the factors of the equation

\(∴s=0\)

\(B=\frac{7}{25}\)

\(∴s=-25\)

\(A=-\frac{7}{625}\)

\(A + C = 0\)

\(∴C=\frac{7}{625}\)

\(-\frac{7}{625s}+\frac{7}{25s^2}+\frac{7}{625(s+25)}=\frac{7}{s^2(s+25}\) ………(equation -13)

Now, the Inverse Laplace Transform of the equation 13 is determined:

\(L^{-1}\{F(s)\}=L^{-1}\bigg\{\frac{7}{25s^2}-\frac{7}{625(s+25)}\bigg\}\)

\(=L^{-1}\bigg\{\frac{7}{25s^2}\bigg\}-L^{-1}\bigg\{\frac{5}{625(s+25)}\bigg\}\)

\(=\frac{7}{25}×L^{-1}\bigg\{\frac{1}{s^2}\bigg\}-\frac{7}{625}×L^{-1}\bigg\{\frac{1}{(s+25}\bigg\}\)………(equation 14)

Using the generalized solution of Inverse Laplace Transform as stated in the Background Information:

\(L^{-1}\bigg\{\frac{1}{s^2}\bigg\}=t\)

\(L^{-1}\bigg\{\frac{1}{(s+25)}\bigg\}=e^{-25t}\)

Putting the values of the Inverse Laplace’s Transform in the equation (14):

\(L^{-1}\{F(s)\}=\frac{7}{25}×t-\frac{7}{625}×e^{-25t}\)

\(⟹x=\frac{7}{25}\bigg(t-\frac{e^{-25t}}{625}\bigg)\)………equation 15)

\(L^{-1}\{F(s)\}\)* = x because it was assumed that:*

*L *{*x*}* = F(s)*

In Figure 3, the solution of the differential equation using Laplace’s method has been plotted. The graph shows the output response of the rice cooker with respect to time as per equation (15).

Figure 3 shows the change of output response of Rice Cooker compared to its operation time (in seconds). Thus, time or independent variable is plotted along X – Axis and output response of the rice cooker or dependent variable is plotted along Y – Axis.

linear increasing trend has been observed. As the time of operation in terms of seconds increases from 0.00 seconds to 1.00 seconds, the magnitude of output response increases from -0.000448 to 0.28s.

The obtained equation of trendline is shown below:

y = 0.280448x - 0.000448

Here, *y *represents the output response of the Rice Cooker and *x *represents the time measured in seconds.

To find the output response of the device at initial condition, i.e., when time is zero is shown below:

y = 0.280448×0 - 0.000448

y = - 0.000448

Therefore, the output response of the device before the device is even started is equal to -0.000448.

Secondly, the time instant at which there is no output response is given below:

0 = 0.280448x - 0.000448

\(⟹x=\frac{0.000448}{0.280448}\)

⟹ x = 0.0016 s

Therefore, at time (t = 0.0016 s), the output response of the bread toaster is zero.

There are no major data points which are lying on either side of the trendline. By observation, most of the data points are found to lie on the trendline which strengthens the correlation as the number of outliers turns zero. Moreover, the strength of the correlation is explained by a high value of regression coefficient (=1).

**Determination of processing speed:**

\(x(t)=x=\frac{7}{25}\bigg(t-\frac{e^{-25t}}{625}\bigg)\)

Differentiating both sides with respect to time, the resultant equation may be obtained as:

\(\frac{dx(t)}{dt}=\frac{d}{dt}\bigg[\frac{7}{25}\bigg(t-\frac{e^{-25t}}{625}\bigg)\bigg]\)

\(⟹\frac{dx(t)}{dt}=\frac{7}{25}×\frac{d}{dt}\bigg(t-\frac{e^{-25t}}{625}\bigg)\)

\(=>\frac{dx(t)}{dt}=\frac{7}{25}×\bigg[\frac{d}{dt}(t)-\frac{d}{dt}\bigg(\frac{e^{-25t}}{625}\bigg)\bigg]\)

\(=>\frac{dx(t)}{dt}=\frac{7}{25}×\bigg[1-(-25)×\frac{e^{-25t}}{625}\bigg]\)

\(=>\frac{dx(t)}{dt}=\frac{7}{25}×\bigg[1+\frac{e^{-25t}}{25}\bigg]\) ………(equation - 16)

In Figure 4, the solution of the differential equation using the Laplace’s method has been shown. The graph represents the output response of the rice cooker with respect to time as per equation (16).

Figure 4 shows the change of ‘processing speed’ (in s^{-1}) of Rice Cooker compared to its ‘operation time’ (in seconds). Thus, time or independent variable is plotted along X – Axis and processing speed of the rice cooker or dependent variable is plotted along Y – Axis.

A decreasing trend, differing as a polynomial function of sixth order is observed. As the time of operation in terms of seconds increases from 0.00 seconds to 1.00 seconds, initially, the processing speed decreases exponentially between time 0.00 seconds and 0.10 seconds. However, the processing speed becomes constant (approximately) and equal to 0.28 s^{-1} till the last time instant (t = 1.00 s).

The obtained equation of trendline is shown below:

y = 1.0316*x*^{6 }- 3.5442*x*^{5 }+ 4.7811*x*^{4 }- 3.2023*x*^{3 }+ 1.1058*x*^{2 }- 0.1831*x* - 0.2912

Here, *y *represents the processing speed (measured in s^{-1}) of the Rice Cooker and *x *represents the time measured in seconds.

There are no major data points which are lying on either side of the trendline. By observation, most of the data points are found to lie on the trendline which strengthens the correlation as the number of outliers turns zero. Moreover, the strength of the correlation is explained by a high value of regression coefficient (= 1).

**Research Question: Determining the processing speed of a Rice Cooker by the investigation of the transfer function (output response with respect to input response) of a Bread Toaster obtained in two different methods (Integrating Factor Method and Laplace’s Method).**

The output response of a rice cooker is increasing in nature with respect to ‘time’(measured in seconds) and has a linear trend line. Thus, there is an increase in the speed of processing with time. Moreover, the rice cooker’s speed of processing is decreasing in nature at the initial stage but it can be seen that it eventually becomes constant after time instant (t = 0.1s).

Nevertheless, to find the solution of differential equations of output response derived from the transfer function, the Laplace's Method has been found to be more suitable and dependable. The reason for this is that the integrating factor method which has been used to solve the differential equation has a variable which is the constant of integration which cannot be found without experimental observation.

A graphical representation has also been shown which further depicts that the integrating factor method results in sufficient error as the nature of the speed of processing that has been obtained is a constant function and has a linear trend line.

However, the trend line of speed of processing that has been found by the use of Laplace’s Method is a sixth-degree polynomial function. The method of Laplace does not result in any error as it involves definite integration. Hence, there is an accurate calculation with no error in the solution as no assumption has been required.

- The solution of differential equation of output response of Rice Cooker using Method of Integrating Factor is:

\(x=\frac{1}{e^{25t}}\bigg[\frac{7e^{25t}}{25}\bigg[t-\frac{1}{25}\bigg]+c\bigg]\)

- The solution of differential equation of output response of Rice Cooker using Laplace’s Method is:

\(x=\frac{7}{25}\bigg(t-\frac{e^{-25t}}{625}\bigg)\)

- Graphically explored trendline of the solution of differential equation of output response of Rice Cooker using Method of Integrating Factor:

y = 0.28x - 0.0112

- Graphically explored trendline of the solution of differential equation of output response of Rice Cooker using Laplace’s Method:

y = 0.280448x - 0.000448

- The equation of processing speed obtained from the solution of differential equation of output response of Rice Cooker using Method of Integrating Factor is:

Processing Speed = 0.28

- The equation of processing speed obtained from the solution of differential equation of output response of Rice Cooker using Laplace’s Method is:

Processing Speed = \(\frac{7}{25}×\bigg[1+\frac{e^{-25t}}{25}\bigg]\)

- Graphically explored trendline of the processing speed of the Rice Cooker obtained from the solution of differential equation of output response of Rice Cooker using Method of Integrating Factor:

y = 0.28

- Graphically explored trendline of the processing speed of the Bread Toaster obtained from the solution of differential equation of output response of Bread Toaster using Laplace’s Method:

y = 1.0316*x*^{6 }- 3.5442*x*^{5 }+ 4.7811*x*^{4 }- 3.2023*x*^{3 }+ 1.1058*x*^{2 }-0.1831*x* - 0.2912

- A Graphical representation has been used to understand the equation of the response of output of the rice cooker and of the speed of processing of the equation. This representation is of immense significance and has a major contribution in proving the existence of the equation which has been obtained by using two methods for finding the solution of the differential equations.
- Examining the obstacles of the graph which has been obtained from the equation of the response of output of the device with respect to time and from the equation of speed of processing of the rice cooker with respect to time (measured in seconds) has allowed me to draw the conclusion of a few significant drawbacks of the physical quantities which are related to operation of the rice cooker.
- As all the graphs have the same time range, it has enabled an accurate comparative analysis between the output response and speed of processing of the rice cooker which has been solved using the integrating factor method and the method of Laplace.

- As different sites from the internet have been used to find out the differential equation of output response, there cannot be a verification of the existence of the equation before using them in the exploration. This makes the existence and constancy of the exploration debatable.
- The integrating factor method which has been mentioned in the background information is valid only for the first order differential equations. Though the transfer function of the rice cooker could have been a differential equation of orders greater than 1, it was not investigated whether the exploration is valid for differential equations for an order higher than 1.
- Due to the presence of an integration constant which has been obtained in finding the solution of the differential equation, which was further obtained by using the integrating factor method, the original equation cannot be found. An experimental process could have been used to obtain the original equation but it has not been followed in this exploration.

Method of Integrating Factor is a greatly used approach for evaluating differential equations. It is used to evaluate first order differential equations only; but it can be used to evaluate higher order differential equations too. Thus, another exploration could be done on comparing between the usage of method of integrating factor for two different orders of differential equations. The research question could be framed as: “Exploring the solution of two distinct differential equations obtained as a transfer function (mathematical interpretation of the input and output signal) of first order and second order for a Rice Cooker using the method of integrating factor to determine the usability of application of method of integrating factor for higher order differential equations.”

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- Electrical4U. ‘Transfer Function of Control System | Electrical4U’. https://Www.Electrical4u.Com/, https://www.electrical4u.com/transfer-function/. Accessed 19 June 2021.
- Differential Equations. https://tutorial.math.lamar.edu/classes/de/de.aspx. Accessed 17 June 2021.
- Differential Equations - Euler’s Method. https://tutorial.math.lamar.edu/classes/de/eulersmethod.aspx. Accessed 17 June 2021.
- Binder, Andrew. Integrating Factor Method. 17 Feb. 2012,
- The Laplace Transform Applications.
- Listing of Laplace and Inverse Laplace Transforms. https://math.info/Differential_Equations/Laplace_Transforms_List/. Accessed 19 June 2021.
- ‘Calculus - Finding a General Solution to a Differential Equation, Using the Integration Factor Method’. Mathematics Stack Exchange, https://math.stackexchange.com/questions/1384576/finding-a-general-solution-to-a-differential-equation-using-the-integration-fac. Accessed 17 June 2021.