This presentation explains the basic information about Polynomial Function and Synthetic Division. Examples were given about easy ways to divide polynomial function using synthetic division. It also contains the steps on how to perform the division method of polynomial functions.
Complementary function, particular integral,homogeneous linear functions with constant variables, Euler Cauchy's equation, Legendre's equation, Method of variation of parameters,Simultaneous first order linear differential equation with constant coefficients,
Solutions manual for college algebra graphs and models 5th edition by bittingerKrisWu4678
Solutions Manual for College Algebra Graphs and Models 5th Edition by Bittinger
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This presentation explains the basic information about Polynomial Function and Synthetic Division. Examples were given about easy ways to divide polynomial function using synthetic division. It also contains the steps on how to perform the division method of polynomial functions.
Complementary function, particular integral,homogeneous linear functions with constant variables, Euler Cauchy's equation, Legendre's equation, Method of variation of parameters,Simultaneous first order linear differential equation with constant coefficients,
Solutions manual for college algebra graphs and models 5th edition by bittingerKrisWu4678
Solutions Manual for College Algebra Graphs and Models 5th Edition by Bittinger
Full download at: https://goo.gl/pgyvya
college algebra graphs and models 5th edition pdf
college algebra graphs and models 6th edition pdf
college algebra graphs and models pdf
college algebra graphs and models 5th edition online
college algebra 5th edition beecher penna bittinger pdf
college algebra graphs and models 4th edition pdf
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How to handle Initial Value Problems using numerical techniques?
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How to handle a system of initial value problems using Runge-Kutta method?
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https://wikicourses.wikispaces.com/Topic+Initial+Value+Problems
TEDx Manchester: AI & The Future of WorkVolker Hirsch
TEDx Manchester talk on artificial intelligence (AI) and how the ascent of AI and robotics impacts our future work environments.
The video of the talk is now also available here: https://youtu.be/dRw4d2Si8LA
Numerical integration is the approximate computation of an integral using numerical techniques. The numerical computation of an integral is sometimes called quadrature. ... A generalization of the trapezoidal rule is Romberg integration, which can yield accurate results for many fewer function evaluations.
Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers), structure, space, and change. There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics
This material is a part of PGPSE / CSE study material for the students of PGPSE / CSE students. PGPSE is a free online programme for all those who want to be social entrepreneurs / entrepreneurs
Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
Explore the innovative world of trenchless pipe repair with our comprehensive guide, "The Benefits and Techniques of Trenchless Pipe Repair." This document delves into the modern methods of repairing underground pipes without the need for extensive excavation, highlighting the numerous advantages and the latest techniques used in the industry.
Learn about the cost savings, reduced environmental impact, and minimal disruption associated with trenchless technology. Discover detailed explanations of popular techniques such as pipe bursting, cured-in-place pipe (CIPP) lining, and directional drilling. Understand how these methods can be applied to various types of infrastructure, from residential plumbing to large-scale municipal systems.
Ideal for homeowners, contractors, engineers, and anyone interested in modern plumbing solutions, this guide provides valuable insights into why trenchless pipe repair is becoming the preferred choice for pipe rehabilitation. Stay informed about the latest advancements and best practices in the field.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
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In today’s fast-changing business environment, it’s extremely important to be able to respond to client needs in the most effective and timely manner. If your customers wish to see your business online and have instant access to your products or services.
Online Grocery Store is an e-commerce website, which retails various grocery products. This project allows viewing various products available enables registered users to purchase desired products instantly using Paytm, UPI payment processor (Instant Pay) and also can place order by using Cash on Delivery (Pay Later) option. This project provides an easy access to Administrators and Managers to view orders placed using Pay Later and Instant Pay options.
In order to develop an e-commerce website, a number of Technologies must be studied and understood. These include multi-tiered architecture, server and client-side scripting techniques, implementation technologies, programming language (such as PHP, HTML, CSS, JavaScript) and MySQL relational databases. This is a project with the objective to develop a basic website where a consumer is provided with a shopping cart website and also to know about the technologies used to develop such a website.
This document will discuss each of the underlying technologies to create and implement an e- commerce website.
Saudi Arabia stands as a titan in the global energy landscape, renowned for its abundant oil and gas resources. It's the largest exporter of petroleum and holds some of the world's most significant reserves. Let's delve into the top 10 oil and gas projects shaping Saudi Arabia's energy future in 2024.
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A hierarchical digital twin of a Naval DC power system has been developed and experimentally verified. Similar to other state-of-the-art digital twins, this technology creates a digital replica of the physical system executed in real-time or faster, which can modify hardware controls. However, its advantage stems from distributing computational efforts by utilizing a hierarchical structure composed of lower-level digital twin blocks and a higher-level system digital twin. Each digital twin block is associated with a physical subsystem of the hardware and communicates with a singular system digital twin, which creates a system-level response. By extracting information from each level of the hierarchy, power system controls of the hardware were reconfigured autonomously. This hierarchical digital twin development offers several advantages over other digital twins, particularly in the field of naval power systems. The hierarchical structure allows for greater computational efficiency and scalability while the ability to autonomously reconfigure hardware controls offers increased flexibility and responsiveness. The hierarchical decomposition and models utilized were well aligned with the physical twin, as indicated by the maximum deviations between the developed digital twin hierarchy and the hardware.
Numerical Method for UOG mech stu prd by Abdrehman Ahmed
1. Bisection method
1.Using bisection method find the positive root
of the following function F(x)=x3-4x-9
solution; here we can conclude one
root of f(x) lie between 2 & 3 i.e;
f(2)=-9 (-ve) and f(3)=6 (+ve)
If xr represents the root mean of the function
by bisection methods
I. Xr=(2+3)/2=2.5
f(xr)=f(2.5)=-3.375 is –ve
error=(2.5-3)/2.5*100=20%
II. Xr=(2.5+2.75)/2=2.75
f(xr)=f(2.75)=0.796875 is +ve
III. Xr=(2.5+2.75)/2=2.625
f(xr)=f(2.625)=-1.412109 is –ve
Continuous up two required iteration
x F(x)
0 -9
1 -12
2 -9
3 6
2. Secant method
xi+1=xi-[f(xi)(xi-1-xi)]/[f(xi-1)-f(xi)]
1.Use the secant method to estimate the root of
f(x)=x3-x-1 with initial estimates of 1&2 having
three iteration
Solution;
xi-1=1 and f(2)=5
I. X1=2-[f(2)(1-2)]/[f(1)-f(2)]=2-[5(-1)]/[-1-5] = 1.166667
Now xi-1=2 and xi=1.166667
error= [1.166667-2]/1.166667*100= 71.4286%
II. F(2)=5 and f(1.166667)= -0.5787
X2=1.166667-[-0.5787(2-1.166667)]/[5-(-0.5787)] = 1.25311
Now xi-1 = 1.166667 and xi = 1.25311
III. F(1.166667)=-0.5787 and f(1.25311)= -0.28537
X3 = 1.25311-[-0.28537(1.16667-1.25311)]/[-0.5787-(-0.28537)]
X3= 1.3372 f(1.3372) = 0.0538854
3. Lease square regression method
1.Find the equation of straight line which best fits
n = 7 ,
Ẋ = ∑xi/n
Ý= ∑yi/n ,
a1=[n∑xiyi-∑yi]/[n∑xi2-(∑xi)2
ao = Ý-a1Ẋ
therefore the line which fits the given tabular
points is y= ao+a1x
x 10 12 13 16 17 20 25
y 10 22 24 27 29 33 37
Xi Yi XiYi xi2
10 10 100 100
12 22 264 144
13 24 312 169
16 27 432 256
17 29 493 289
20 33 660 400
25 37 925 625
Sum 113 182 3186 1983
4. Fourier approximation
1.The PH in a reactor varies sinusoidally as the couse
of a day use Fourier approximation
Time(h) 0 2 4 5 7 8.5 12 15 20 22 24
PH 7.3 7 7.1 6.4 7.2 8.9 8.8 8.9 7.9 7.9 7
Where f=1/t and t=24hr , f=1/24= 0.042hr-1 ,n = 11
Wo=2πf=0.261799388
Ao=∑y/n, A1 = 2/n[∑ycos(wot) ,
B1 = 2/n[∑ysin(wot)
The equation will be y= Ao+A1cos(wot)+B1sin(wot)
time(t) PH(y) ycos(wot) ysin(wot)
0 7.3 7.3 0
2 7 6.062177825 3.500000002
4 7.1 3.549999995 6.14878037
5 6.4 1.656441882 6.18192529
7 7.4 -1.915260944 7.147851112
8.5 7.2 -4.383082299 5.712144043
12 8.9 -8.9 -2.14508E-08
15 8.8 -6.222539656 -6.222539693
20 8.9 4.450000031 -7.707626076
22 7.9 6.841600707 -3.94999997
24 7 7 3.37429E-08
Sum 83.9 15.43933754 10.81053509
5. Linear interpolation
The general formula for linear interpolation is
𝑓1( 𝑥)−𝑓(𝑋𝑜)
𝑋−𝑋𝑂
=
f(X1)−f(Xo)
𝑋1−𝑋0
from the above we get the
general formula for linear interpolation
f1(x) = 𝑓(𝑥𝑜) +
𝑓( 𝑥1) − 𝑓( 𝑥𝑜)
𝑥1 − 𝑥𝑜
(𝑥 − 𝑥𝑜)
Example find the value of ln(2) if ln(1)=0 and
ln(4)=1.386294 using linear interpolation
Solution 1st we make a table
Given f(xo)=0, xo=1
f(x1)=1.386294, x1=4
f1(x)=?, x=2
By using linear interpolation formula
f1(x) = 𝑓(𝑥𝑜) +
𝑓( 𝑥1) − 𝑓( 𝑥𝑜)
𝑥1 − 𝑥𝑜
(𝑥 − 𝑥𝑜)
f1(x) = 0 +
1.386294 − 0
4 − 1
(2 − 1)
f1(x) =
1.386294
3
= 0.4621 𝑎𝑛𝑠𝑤𝑒𝑟
x ln(x)
1 0
2 ?
4 1.386294
6. The gauss seidel method
Rule
|a11|>|a12|+|a13|,
|a22|>|a21|+|a23|, and
|a33|>|a31|+|a32|
Iteration 1, y = z =0, xi=?,
x = xi, z = 0 , yi = ?
x = xi, y = yi, zi = ?
Iteration 2,
using yi & zi we get xii = ?
xii & yi we get yii = ?
xii & yii we get zii = ?
continue this to the required iteration
a11 a12 a13
a21 a22 a23
a31 a32 a33
e
f
g
7. The naïve gauss elimination method
Example; given 20x+y+4z=25----------------eq(1)
8x+13y+2z=23--------------eq(2)
4x-11y+21z=14-------------eq(3)
with the value of x=y=z=1
Solution: by multiplying eq(1) by 8/20 and subtract
from eq(2) and also multiply eq(1) by 4/20 and
subtract from eq(3) we get the following set of equations
20x+y+4z=25--------------------eq(1)
12.6y+0.4z=13------------------eq(4)
-11.2y+20.2z=9-------------------eq(5)
by proceeding to eliminate we find a single variable z by multiplying
eq(4) by -11.2/12.6 and subtract from eq(5) we get
20.5556z=20.5556----------------eq(6)
finally we rearrange the equation and written as
20x+y+4z=25 -----------------eq(1)
12.6y+0.4z=13 ----------------eq(4)
20.5556z=20.5556-----eq(6)
now by using back substitution we get the unknowns x,y&z
from eq(6) we get z=1 ,
from eq(4) we get y=1,
from eq(1) we get x=1
9. Matrix inverse using LU decomposition
We find the LU decomposition of the given matrix as the
above form and then we find the inverse of the matrix
Procedure one; [L] =
to find the value of d1,d2,d3 by using forward
substitution
d1+0+0=1 we get d1
F21d1+d2+0=0 we get d2
F31d1+F32d2+d3=0 we get d3
the vector can be then used as the right hand side for
the upper triangular matrix
[U] =
let us find the value of x1,x2 and x3 using backward
substitution
0+0+a33x3=d3 from this we get x3
0+a22x2+a23x3=d2 from this we get x2
a11x1+a12x2+a13x3=d1 from this we get x1
d1
d2
d3
1
0
0
d1
d2
d3
x1
x2
x3
10. x1,x2 and x3 are the first column of the inverse matrix
As like as procedure one we get all the columns of the
inverse matrix by only changing
To and using in procedure 2 and 3
respectively
finally the inverse of the matrix we get is
[A]-1 =
1
0
0
0
1
0
0
0
1
X1 X1 X1
X2 X2 X2
X3 X3 X3
11. Trapezoidal rule
Examples;
∫0
6 1
1 𝑥
dx given strip 6,
Solution; given a = 0 and b = 6
Interval = b-a = 6 and
width of strip = h = interval/no of strip = 6/6 = 1
integration =
h/2 [ (sum of the first and last ordinate)
+ 2(sum of remaining ordinates)]
integration=
½[(1+0.142857)+2(0.5+0.3333+0.25+0.2+0.166667)]
integration = 2.021429
actual value = 1.945910 integrate the above equation
therefore the error =
[(actual value – new value)/actual value]*100%
Error = =3.880%
x 0 1 2 3 4 5 6
f(x) 1 0.5 0.333333 0.25 0.2 0.166667 0.142857
12. Simpson’s 1/3 rule
To find Simpson’s 1/3 rule we use the following
equation
integration = h/3 [ (sum of the first and last
ordinates) + 2(sum of even ordinates) + 4(sum of
odd ordinates) ]
Example; ∫0
6 1
1 𝑥
dx given strip 6,
Solution; given a = 0 and b = 6
Interval = b-a = 6 and
width of strip = h = interval/no of strip = 6/6 = 1
Integration =
1/3[(1+0.142857)+2(0.3333+0.2)+4(0.5+0.25+0.166667) ]
integration = 1.95873
actual = 1.945910
Error=[(actual value – new value)/actual value]*100%
Error = = 1.3553%
y0 y1 y2 y3 y4 y5 y6
x 0 1 2 3 4 5 6
f(x) 1 0.5 0.333333 0.25 0.2 0.166667 0.142857
13. Romberg integration
Example1; ∫1
2
(𝑥 +
1
)2
dx and actual value is 4.83333
Solution; a=1 and b=2 f(x) = (𝑥 +
1
)2
Step I, f(1) = 4 and f(2) = 6.25
A.n1 = 1 , h1=(b-a)/n1= (2-1)/1 = 1
I1 = h1*[f(1)+f(2)]/2 = 1*[4+6.25]/2
I1 = 5.125
error=[(actual value – new value)/actual value]*100%
Error = 6.0418%
B.n2 =2 , h2 = (b-a )/n2 = 0.5 is width of strip
as a result of h2 = 0.5
we get f(1)=4, f(1.5)=4.6944 and f(2)=6.25
I2 = h2 * [ (f(1)+f(1.5))/2] + h2 * [(f(1.5)+f(2))/2]
I2 = 4.909722
error = [(actual value – new value)/actual value]*100%
Error = 1.58047%
C.n3 = 4, h3 = (b-a)/n3 =0.25 is width of strip
as a result of h3 = 0.25 we get
f(1)=4, f(1.25) , f(1.5)=4.6944, f(1.75), and
f(2)=6.25
15. Euler method
The formula for Euler method is
Yi + = yi + f(xi,yi) h
Examples solve dy/dx = -2x3+12x2-20x+8.5
for x=0, step size = 0.5 up to x = 4
Solution; x=0,0.5,1,1.5,2,2.5,3,3.5,4
dy/dx = -2x3+12x2-20x+8.5 integrate both sides
y = -1/2x4+4x3-10x2+8.5x+c , f(0) = c = 1
y = -1/2x4+4x3-10x2+8.5x+1 true solution
the initial condition at x = 0 is y = 1 thus xi = 0& yi = 1
y(0.5) = y(0) + f(0,1) *step size = 5.25 predicted value
The true solution at x = 0.5 is
y = −0.5(0.5)4 + 4(0.5)3 − 10(0.5)2 + 8.5(0.5) + 1 = 3.21875
Error =[ (tv-pv)/tv]*100% = [3.218-5.25)/3.218]*100%=63.1%
From the above we get xi = 0.5 and yi = 5.25 then
y(1) = yi + f(xi,yi) h = y(0.5) + f(0.5,5.25)*0.5 = 5.875 is
euler value or predicted value
The true solution at x = 1 is
y(1) = −0.5(1)4 + 4(1)3 − 10(1)2 + 8.5(1) + 1 = 3
Error = [ (tv-pv)/tv]*100% = [(3-5.785)/3]*100% = 95.833%
16. From the above we get xi = 1 and yi = 5.875
This is continue up to y(4)
Therefore the table will be formatted as
x Y true Y euler Error
0 1 1
0.5 3.218 5.25 63.1%
1 3 5.875 95.833%
Example ; solve y’ = x + y for y(0) = 1 find value of y at
x = 0,0.2,0.4,0.6,0.8,1
Solution by integrating both sides the problem
becomes y = x2/2 + xy + c, solving for y(0) = 1, c = 1
and substitute y = x2/2 + xy + 1 this is a true solution
h = step size = 0.4-0.2 = 0.2 and xi = 0 & yi =1
y(0.2) = y(0) + f(0,1)h = 1.2 is Euler value
The true solution at x = 0.2 is
y(0.2) = (0.2)2/2 + 0.2y +1
y = 0.2 + 0.2y + 1 = 1.5
Error = [(tv – euler value)/tv]*100% = 20%
from the above we get xi = 0.2 and yi = 1.2 it is
continuous up to the given x value y(1)
17. Leibmann method
The general formula for leibmann method is
Ti,jnew = ለTi,jnew + (1-ለ)Ti,jold
Examples solve the following problems with leibmann
method with ለ = 1.25 and Ea(error) < 1% for 1st element
100oc
750c 500c
00C
1st iteration
T1,1 = [right + left + top + bottom] / 4 = [R+L+T+B]/4
T1,1 = [0+75+0+0] / 4 = 18.75
T1,1new = 1.25*18.75+(1-1.25) *0= 23.44
Error=[(tv-pv)/tv]*100% =[(23.4-0)/23.4]*100%=0%
From the 1st iteration we use Ti,jold = 0& error = 0%
It is continuous up to T3,3new
(1,3) (2,3) (3,3)
(1,2) (3,2)
(1,1) (2,1) (3,1)
(2,2)
18. 2nd iteration
T1,1 =[ R+L+T+B]/4 = [7.324+75+30.37+0]/4 = 28.26
T1,1new = 1.25*28.26+(1-1.25)*23.44 = 29.47
Error=[(current iteration–previous iteration)/ci]*100
Error = [(29.47-23.47)/29.47]*100% = 20.5%
At the 2nd iteration we use the formula
Ti,jnew = ለTi,jnew + (1-ለ)Ti,jold
Where
Ti,jnew = the value we get in the second
iteration of Ti,j
Ti,jold = is the value we get in the first iteration of
Ti,j
Then it is continuous up to T3,3new
19. The table then set us the following form
element 1st iteration 2nd iteration
T 0c Error % T 0c Error %
T1,1 0 100 29.47 20.5
T2,1 0 100 16.64 56.2
T3,1 0 100 24.1 25.7
T1,2 0 100 48.26 36.9
T2,2 0 100 42.46 71.99
T3,2 0 100 52.64 52.7
T1,3 0 100 75.05 9.5
T2,3 0 100 75.4 26.9
T3,3 0 100 68.9 4.3
20. The fourth order Ruge Kutta method
The formula for the fourth order ruge kutta method is
yi+1 = yi + ɸ(xi,yi,h)*h
ɸ = a1k1+a2k2+a3k3+…+ankn
yi+1 = yi + 1/6 [k1+2k2+2k3+k4]*h
where
k1 = f(xi,yi)
k2 = f(xi+h/2, yi+k1*h/2)
k3 = f(xi+h/2,yi+k2*h/2)
k4 = f(xi+h, yi+k3*h)
Examples
1, solve f(x,y)=-2x3+12x2-20x+8.5,with h=0.5 & y(0)=1
solution from the given equation at xi=0 , yi=1
f(x) = )=-2x3+12x2-20x+8.5 then we find the value of k
k1 = 8.5, k2 = 4.21875, k3 =4.21875 and k4 = 1.25
y(0.5) = y(0) + 1/6 [k1+2k2+2k3+k4]*h
y(0.5) = 3.4875
21. 2, f(x,y)=4e0.8x -0.5y, h = 0.5 with y(0) = 2 from x=0 to x=0.5
Solution
find the value of 1st slope xi=0, yi=2, f(x)= 4e0.8x -0.5y
K1 = f(xi,yi) = f(0,2) = 3
This value is used to compute the value of ‘y’ and slope at
mid-point y(0.25) = 2 + 3(0.25) = 2.75, xi=0 and yi=2
k2 = f(xi+h/2, yi+k1*h/2) = f(0.25,2.75)=3.51
This slope in form is used to compute another slope of
mid-point
y(0.25) = 2 + 3.510611(0.25) = 2.877653 ,xi=0 and yi=2
k3 = f(xi+h/2,yi+k2*h/2) = f(0.25,2.8776)=3.412
This value used to compute the value of y and slope at
the end of the interval
y(0.5) = 2 + 3.071785(0.5) = 3.723392, xi=0 & yi=2
k4 = f(xi+h, yi+k3*h)=f(0.5,3.7233)= 41.056
Let’s find an average slope
yi+1 = yi + 1/6 [k1+2k2+2k3+k4]*h
yi+1 = 3.7515
22. 3, find the value of y at x=0.1,0.2 y’=-y, y(0)=1,h=0.1
f(x,y) = -y
Solution let’s find k’s where xi=0.1 and yi=1
k1 = f(xi,yi) = f(0.1,1) = -1
k2 = f(xi+h/2, yi+k1*h/2)= f(0.15,0.95) = -0.95
k3 = f(xi+h/2,yi+k2*h/2)= f(0.15,0.9525)=-0.9525
k4 = f(xi+h, yi+k3*h) = f(0.2,0.9047) = -0.9047
then yi+1 = yi + 1/6 [k1+2k2+2k3+k4]*h
yi+1 = 0.9048374
then to find another value of y put yi=0.9048374, and
use xi = 0.2, h=0.1, we get the same individual slope of
k1= -1
k2 = -0.95
k3 = -0.9525 and
k4 = -0.9047 hence
yi+1 = yi + 1/6 [k1+2k2+2k3+k4]*h
yi+1 = 0.8097 note that for other questions the
values of k’s may be difference