First order linear differential equation

First order differential equations: A first order differential equation is a relation dy/dx = f(x, y).....(1)in which f(x, y) is a function of two variables defined on a region in the xy-plane. A solution of the equation (1) is a differential function y = y(x) defined on an interval of x-values such that d/dx y(x) = f(x, y(x)) on that interval.
The initial condition that y(x0) = y0 amounts to requiring the solution curve y = y(x) to pass through the point (x0, y0).The general form of a first order and first degree differential equation is f(x, y, dy/dx) =0....(i).we know that the tangent of the direction of a curve in Cartesian rectangular coordinates at any point is given by dy/dx .

so the equation in (i)  can be known as the equation which establish the relationship between the coordinates of a point and the slope of the tangent ie dy/dx to the integral curve at a point .Solving the differential equation  given by (i) means finding those curves for which the direction of tangent at each point coincides with the direction of the field.All the curves represented by the general solution when takes together will give the locus of the differential equation .Since there is one arbitrary constant in the general  solution of the equation of the first order. The locus of the equation can be said to be made up of single infinity of curve.

First Order Linear Differential Equation A first order differential equation that can be written in the form  dy/dx + P(x) y = Q(x),where P and Q are functions of x, is a linear first order equation. The above equation is the standard form.  Let us understand First Order Differential Equation Solver using some example. Suppose we have the equation dy/dx = 1 – y/x is a first order differential equation in which f(x, y) = 1 – (y/x).
now let us take one more example show that the function y = 1/x + x/2 is a solution of the initial value problem dy/dx = 1 – y/x,y(2) = 3/2 . The given function satisfies the initial condition because y(2) = (1/x + x/2)x = 2= ½ + 2/2 = 3/2To show that it satisfies the differential equation, we show that the two sides of the equation agree when we substitute (1/x) + (x/2) for y.On the left: dy/dx = d/dx(1/x + x/2) = -1/x^2 + ½.On the right: 1 – y/x = 1 – 1/x(1/x + x/2) = 1 – 1/x^2 – ½ = -1/x^2 + ½.The function y = (1/x) + (x/2) satisfies both the differential equation and the initial condition, which is what we needed to show.

What is a Mode in Math?

Math teacher, Ms Grace assigned the following number of problems for homework on 8 different days; 10, 12, 8, 10, 7, 10, 8, 10. Let us arrange the data in the increasing order, 7,8,8,10,10,10,10,12. What do you observe in the data? 10 is the number which is seen most number of times. This number 10, which occurs most often, is called the Mode in Math or Mode Math. So, Mode Definition Math would be, the Mode of a set of data is the value in the set that occurs most often (most number of times)

By definition, Statistics Mode or statistical mode is the value that occurs most frequently in the data set. For example, what would be the mode of the data set, 12, 11, 13, 9, 11, 8, 6, 11. Let us first arrange the data in the ascending order. That gives us, 6, 8, 9, 11, 11, 11, 12, 13. As you can see the value or the number 11, occurs most number of times than any other number in the data and hence 11 is the mode of the given data set.

To learn the steps involved in finding mode of a given set, let us consider an example. Let us assume that there is a basket ball match going on and the scores of the game are as listed below. Let us determine the mode of the scores.

Scores: 19, 6, 3, 22, 19, 9

Step1: list in the scores in the ascending order
3, 6, 9, 19, 19, 22
Step2: identify the number which is occurs most often, 19
  Hence,  19 is the mode of the score

So, the steps involved in finding mode are, first we need to order the list in ascending order and then identify the number or value which occurs most often that will give us mode.

Note: If there is no number which occurs most often, then we can say that the given data has no mode
On a cold winter day in December, the temperature of 7 cities in North America is recorded in Fahrenheit. How do we find the mode of these temperatures?  Temperatures recorded:  -9,-12, 0, -6,      - 10, -3, 5
To find the mean, the first step would be to order the given temperatures in ascending order and then identify the temperature which occurs most often
-12, -10, -9, -6,-3, 0, 5
From the above ordered list, we can see that there is no value or number which occurs most often. So, we can conclude that there is no mode for the given temperatures.

This article gives basic information about Online Statistics Tutoring. Next article will cover more Statistics concept and its problems and many more. Please share your comments.

Definition of Absolute Value

Definition of Absolute Value
The absolute value is defined as the distance of ‘a’ from zero and is denoted by the symbol |a|. The absolute value only tells how far from zero and not in which direction is the location of a.

From the above figure, the absolute value of |3| = 3 since it is 3 units from the right side of the zero, and |-3|= 3 since it is 3 unit from the left side of the Zero. So the absolute unit should not have the negative sign, and is always positive or zero.

Absolute value equation

To solve absolute value equations, first split the equation into two equations. Then solve the equation to get the solution of absolute value equation.
Example 1:
|x| = 7
X = 7 x = -7
The solutions are {7, -7}
Example2:
|3x-4| = 5
3x-4  = 5 3x-4 = -5
3x = 9 3x = -1
X = 3 x = -1/3
The solutions are {1/3, 3}


Absolute Value Inequality
To solve absolute value inequalities first isolate the one side on the inequality symbol. Then write the first equation with out absolute sign and solve the inequalities and write second equation without absolute sign, reverse the inequalities and then solve the problem. The absolute value should be greater than any negative number and it should not be less than a negative number, sincethe absolute value should be always positive.
Example1:  (greater than)
|x-20| > 5
x-20> 5 X-20 < -5
x> 25 x < 15
The solution is 15 > x > 25
Example2: (less then or equal to)
|X-3| = 4
X-3 = 4 X-3 = -4
X = 7 X = -1
The solution is -1 = x = 4

Absolute value of a complex number
The definition for absolute value ofa real number is not directly generalized for definition for complex numbers, since the complex numbers are not ordered. But the geometric interpretation of the absolute value of a real number is as its distance from ‘0’ to be generalized.

The definition for absolute value ofa complex number is, it is a distance in the complex plane from the origin ‘0’ by using of Pythagorean Theorem. The absolute value of the difference of the complex number is equal to the difference between them.
The definition for absolute value ofa complex number is,
Z = x + iy
Where,
Z - Absolute value or modulus
x, y -  Real numbers
|z| = v(x^2+y^2 )
In above equation, the absolute value of a real number is x, when the complex part becomes 0.
In polar form the complex number z is expressed like z = re i?
If the absolute value is (r = 0, ?-real)
|z| = r
So, the absolute value of a complex number in the complex analogue equation is,

The properties of absolute value of a complex number are as same as absolute values of a real numbers.

Give example how to solve the absolute value equations and inequalities
Example 1: absolute value equation
|x - 7| = |2x-2|
Write in to two equations with out absolute symbol.
X-7 = 2x-2 x-7 = -(2x-2)
X-2x = -2+7 x+2x = 7 + 2
-x = 5 3x = 9
X = -5 x = 9
The solutions are { 9, -5}
Example 2: Absolute value inequalities
|2x-3| > 5
Write in to two inequalities with out absolute symbol.

2x-3 > 5 2x-3 < - 5
2x > 8 2x< -2
X < 4 x < -1
The solution is -1 > x < 4

Equation of a line which is passing through two points

Question :-

Find the equation of the line which is passing through two points (-3,7)(5,-1)

Answer:-

We have to use the point formula to find the equation of the line,this is much similar to midpoint formula

y-y1     x-x1
------ = ------
y2-y1     x2-x1

We have 2 points

( -3 , 7 )  and ( 5 , -1 )
x1  y1          x2  y2

So the equation is

y-7      x-(-3)
------ = ------
-1-7      5-(-3)

y-7      x+3
------ = ------
-8        8

We can further simplify it by cross multiplication.which is a part of indices maths

similarly we can find all points having an x-coordinate of 2 whose distance from the point 2 1 is 5

Statistics: Mean

Mean Math Or Mean Statistics
In Mathematics in the branch of Statistics, the expression for the mathematical mean of a statistical distribution is the mathematical average of all the terms in the data. To calculate this, we add up the values of the terms given and divide the sum by the number of terms in the data. This expression is also called the Arithmetic Mean.
Example: Find the Arithmetic Mean of the following data 5, 5,10,10,15,15,20,30
Solution: Arithmetic Mean = Regular average = sum of the values of the terms/number of terms
Sum of the values of the terms =5+5+10+10+15+15+20+30=110
Number of terms = 8
Mean = 110/8 = 13.75
Sample Mean
The sample mean in statistics branch of Mathematics is the sum of all observed outcomes from the sample divided by the total number of events. It is denoted by the symbol x with a (bar) above it. The formula used to compute the sample mean is as follows:
X (bar)= (1/n) (x1+x2+x3………xn)
 If we consider the sampling of some non-indigenous weed in a land of five acres in Springfield and came up with the counts of this weeds in that area as 38, 56, 84,105,116
We calculate sample mean for the above sampling by adding the weed counts and divide by the number of samples, 5
(38+56+84+105+116)/5= 79.8
So, we get the sample mean of non-indigenous weeds = 79.8
Weighted Mean
While computing Arithmetic Mean all the terms or values have equal importance. But at times we come across some situations where all the terms or values do not have equal importance. For instance, when we compute the average number of marks per subject, as per the different subjects like English, Science, Mathematics, Social Sciences; each subject has different levels of importance.
So, weighted mean is the arithmetic mean calculated by considering the relative importance (weight) of each term.
Each item is assigned a weight in proportion to its relative importance
Formula to compute Weighted Mean: xw (bar) = sigma(wx)/sigma(w)
(here x =value of the items  w= weight of the item)
Example:  A student scored 50, 70, 80 in English, Science and Math respectively and assume the weights of each subject to be 4,5,6  respectively. Find the weighted arithmetic mean of each subject.
Solution: let us tabulate the given data
Subjects              Marks obtained          weight              wx
English                50                          4                  150
Science                       60                          5                  300
Math    80                          6                  480
         Sigma(w)=15  sigma(wx)=930
Arithmetic Weighted mean = 930/15= 62 marks/subject
Short-cut Method
A short-cut method of calculating the arithmetic mean is based on the property of arithmetic average. In this method the deviations (D) of the items from an assumed mean are first calculated and then multiplied with their respective frequencies (f). Then, the total of these products [sigma(fD)] is divided by the total frequencies [sigma(f)]and added to the assumed mean(A). The figure we get is the actual arithmetic average or the Arithmetic Mean.
Formula used in the short-cut method of calculating the arithmetic mean:
X(bar) = A + sigma(fD)/sigma(f)

Operations on Decimals

Decimals:
Numbers representing fractions without having any numerator or denominator is called a decimal number.
Example: 0.567, 0.2, 0.49
Decimal Chart:
The value of a number in a decimal is decided by its place value.
0.345
3 represents three tenths = 3/10
4 represents four hundredths = 4/100
5 represents five thousandths = 5/1000

In engineering calculations, fractions are used extensively. As we know that, we can always express a fraction in terms of its corresponding decimal number. We have a ready reckoner that gives us corresponding converted values of a fraction.
1/64   = .015625 1/32   = .03125 3/64  = .046875 1/16 = .0625
5/64   = .078125 3/32   = .09375 7/64  = .109375 1/8  = .125
9/64   = .140625 5/32   = .15625 11/64 = .171875 3/16 = .1875
13/64   = .203125 7/32   = .21875 15/64 = .234375 1/4  = .25
17/64   = .265625 9/32   = .28125 19/64 = .296875 5/16 = .3125
21/64   = .328125 11/32  = .34375 23/64 = .359375 3/8  = .375
25/64   = .390625 13/32  = .40625 27/64 = .421875 7/16 = .4375
29/64   = .453125 15/32  = .46875 31/64 = .484375 1/2  = .5
33/64   = .515625 17/32  = .53125 35/64 = .546875 9/16 = .5625
37/64   = .578125 19/32  = .59375 39/64 = .609375 5/8  = .625
41/64   = .640625 21/32  = .65625 43/64 = .671875 11/16= .6875
45/64   = .703125 23/32  = .71875 47/64 = .734375 3/4  = .75
49/64   = .765625 25/32  = .78125 51/64 = .796875 13/16= .8125
53/64   = .828125 27/32  = .84375 55/64 = .859375 7/8  = .875
57/64   = .890625 29/32  = .90625 59/64 = .921875 15/16= .9375
61/64   = .953125 31/32  = .96875 63/64 = .984375

Multiplying decimals
In multiplication of decimal numbers, we follow two steps:
Step 1: Multiply the numbers without taking into consideration of decimal points
Step 2: Place the decimal point starting from right and moving as many digits as the sum of the number of digits in decimal part in both numbers towards left.
Example:


Dividing decimals:
We have two steps to follow.
Step 1: Convert both numerator and denominator into fractions
Step 2:  Follow the rules of division of fractions.
Example: 14.4 ?0.12
Step 1: 14.4 = 144/10 and 0.12 = 12/100
Step 2: 144/ 10 ? 12/100 = 144/10 ?100/12 = 120
Subtracting decimals:
To subtract decimals one from the other we follow
Step 1: Write the numbers one below the other with decimal point coinciding
Step 2: Add zeros to make length of the numbers same
Step 3: Follow the laws of subtraction to subtract
Example:
0.56- 0.287
0.560
0.287
-------
0.213
Terminating decimals:
The word terminate means end. A decimal number with definite number of digits (ending) is called terminating decimal.
Example: 0.495. There are three digits after the decimal point.
All terminating decimals can be represented as a rational number.
Dewey Decimal System:
In public libraries, books are catalogued according to the subjects. Dewey Decimal System (DDC) is the most widely used system of classification for organizing books in libraries. In this system, integer part of the number gives us the specific subject and decimal part gives us the specific part of the subject.
For example:
In DDC, 512.12 represents a book on Algebra, 815.4 represents a book on 16th century Italian poetry.

Exponents and their rules

Fractional exponent is also called as a rational exponent which is in the form of a fraction like ½ (square root), ⅓ (cube root), ¼ (fourth root) etc.

A fractional exponent with 1/n is the n-th root of the given base.

(a)^ ⅓ = ³√a

A fractional exponent with m/n, we need to take the m-th power of the base and then find the n-th root or vice versa

(a) ^⅔ = ³√a²

The basic exponential rules are as follows: 

Exponents and tLaw of exponents:

1. Any number except zero when raised to zero equals 1
2. Any number raised to 1 equals itself
3. A number with negative power equals its reciprocal with a positive power
4. When we divide terms with same bases, the powers are subtracted
5. When we multiply terms with same bases, the powers are added
6. When a term is raised to a power with a whole power, the powers are multiplied.
7. When product of terms are raised to a power, each term is raised to that power

Subtracting exponents:

When we divide terms with same bases, the exponents are subtracted

The basic exponential rules are as follows: 

For example,

= 3³

Exponent rules:

1. When we multiply terms with same bases, the powers are added

2. When we divide terms with same bases, the powers are subtracted

3. Any number except zero when raised to zero equals 1

4. Any number raised to 1 equals itself

5. When a term is raised to a power with a whole power, the powers are multiplied.

6. When product of terms are raised to a power, each term is raised to that power

7. A number with negative power equals its reciprocal with a positive power