Solving Initial Value Problem

Introduction for solving the initial value problem:

Initial value problem is the part of differential equation in mathematics. Initial value problem is an ordinary differential equation joint with particular value, called the initial condition, of the strange function at a known point in the area of the result. In this article we shall discuss  how to solve these initial value problems in the differential equation Having problem with Integral Calculator with Steps keep reading my upcoming posts, i will try to help you.

Example Problems on Solving the Initial Value Problem:

Example 1:

Solving the differential equation `(dy/dx)=4y` using initial condition `y(1)=2`

Solution:

Given differential equation is `(dy/dx)=4y` , initial condition is `y(1)=2`

Step 1:

4y = 0 when y = 0 Thus the constant y(x) = 0 is a particular solution.

Step 2:

Separate the variables on both sides,

`(dy)/y=4 dx`

Step 3:

Take the integral on both sides,

`ln y=4x+B`

step 4:

Take ln on both sides,

`y=e^B e^4x`

`y=Ce^4x`

where `C = eB` if y > 0, and `C = -eB` if y < 0.

Step 5:

General solution: `y(x)=Ce^(4x)` where C is any constant (C can be 0 from Step 1).

Step 6:

Substitute 1 for x and 2 for y

`2=Ce^4,`

`C=2e^(-4)`

The final solution is `y(x)=2e^(-4) e^(4x) =2e^(-(4-4x))`

Example 2:

Solve the initial value problem

Solution:

`y^'' -3y^' +2y=0` , `y(0)=1` ,  `y^' (0)=-1`

First we find the general solution to the differential equation. We can find the general solution to the homogeneous equation by guessing a solution in the form `y(t)=e^(rt)`

`y^'' -3y^' +2y=0`

`(r^2 -3r +2) e^(rt)`

`(r-1)(r-2) e^(rt)`

if we choose r = 1,2 then we find the general solution `y_c (t)=C_1 e^(-t)+C_2 e^(-2t)` for the homogeneous equation.

Now we find the solution corresponding to the initial conditions. We have the derivatives

`y_c (t)=C_1 e^(-t)+C_2 e^(-2t)`

`y(0)=C_1+C_2`

`y^'(t)=C_1 e^(-t)+C_2 e^(-2t)`

`y^'(0)=C_1-C_2`

We see that `C_1+C_2=1 and C_1-C_2=-1`

Solving the above equations we get `C_1=0` and `C_2=1`

Substitute these values we get the solution

`y^' (t)=e^(-t)`


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Practice Problems on Solving Initial Value Problems:

Problem 1:

Solve the initial value problem

`y^'' +3y^' +2=0` , `y(0)=1` , `y^' (0)=-1`

Answer: `y^(t)=e^(-t)`

Problem 2:

Solve the initial value problem

`y^'' +7y^' +12=0` , `y(0)=1` , `y^' (0)=-1`

Answer: `y^(t)=e^(-3t)`

Pi and Radius Math Examples

Introduction to pi and radius math examples:

Pi is the important symbol in math. Radius is the measure that is mainly used in circles.  Pi and radius are the terms that are mostly used in sphere, cone, and cylinder.  Because these figures are the combination of circle shapes.  In this article we have to discuss about “Pi and radius math examples”.

Radius: radius is the distance between the center  to the point on the circle shapes.

pi: it is constant term, the value is 3.14 or `22/7 `

Brief Description of Pi and Radius Math Examples
Example 1:

Find the area and circumference of the circle whose radius is equal to 7 cm.

Solution:

Given radius is equal to 7cm.

Area:

Area of the circle = p r^2 square units

Pi is represented by the symbol p = `(22)/(7)` or 3.14 and r= 7 cm

Therefore area =  `(22)/(7)` x 7 x 7

That is A= 154 square cm

Perimeter:

Perimeter of the circle = 2 p r square units

Therefore area =2x  `(22)/(7)` x 7

That is A= 44 cm

Example 2:

Find the area and volume of the sphere whose radius is equal to 7 cm.

Solution:

Given radius is equal to 7cm.

Area:

Area of the Sphere = 4 p r^2 square units

Pi is represented by the symbol p = `(22)/(7)` or 3.14 and r= 7 cm

Therefore area = 4 x  `(22)/(7)` x 7 x 7

That is A= 616 square cm

Volume:

Volume of the Sphere = `(4)/(3)` p r 3Cubic units

Therefore Volume =`(4)/(3)` x  `(22)/(7)` x 73

That is V= 1437.3 Cubic cm

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Example 3:

Find the area and volume of the cone whose radius is equal to 7 cm and slant height is equal to 10 cm and height is equal to 6 cm.

Solution:

Given radius is equal to 7 cm and slant height is equal to 10 cm and height is 6 cm.

Area:

Area of the Cone = p r (l + r) square units

Pi is represented by the symbol p =  `(22)/(7)` or 3.14 and r= 7 cm and l= 10 cm

Therefore area = `(22)/(7)` x 7 x (10+7)

That is A= 374 square cm

Volume:

Volume of the Cone = `(1)/(3)` p r^2h Cubic units

Therefore Volume =`(1)/(3)`x `(22)/(7)` x 72x6

That is V= 308 Cubic cm

These are the pi and radius problems in math.

Patterns in Math Fourth Grade

Introduction to patterns in math fourth grade:

A patterns, from the French patron, is a type of the theme of recurring events or objects, sometimes referred to as elements of a set. It can be a template or model which can be used with the generate things or parts of a thing. Let we see about the patterns in fourth grade problems like missing terms and sequences. Understanding formula for prime numbers is always challenging for me but thanks to all math help websites to help me out. (Source - Wikipedia )

Problems Based on Patterns in Math Fourth Grade

Example 1 in patterns in math fourth grade:

Find the missed out term in the given numbers 53,51, 49, ____, 45.
The above stated number is in descending order.

The known difference between the number   53-51 = 51 – 49 = 2.

Hence the given numbers are in descending order so we can subtract the numbers 2 from 49 means the answer is 47 .

Example 2 in math patterns in  fourth grade:

Find the missed term in the given numbers 63,61, 59, ____, 55.

The above stated number is in descending order.

The known difference between the number 63-61=61-59 = 2.

Hence the given numbers are in descending order so we can subtract the numbers 2 from 59 means the answer is 47


More Problems in Patterns for Math Fourth Grade

Example 1: 6 36  216  1296

Find the missing terms in the given patterns 6,36,216,1296, ____, _____
Given that:

6,36,216,1296, ____, _____

Solution:

Step 1:   The starting value of the pattern is 6

Step 2:   The next value of the pattern = 6* 6 = 36

Step 3:   The third value can be written as  36 * 6 = 216

Step 4:   The fourth value is 216 * 6 = 1296

Now the left out value can be written as

1296 * 6  = 7776

7776* 6 = 46656

So the correct pattern is 6,36,216,1296, 7776, 46656

The required solution is 6,36,216,1296, 7776, 46656.

Example 2: Find the missing terms in the given patterns  13,169,2197,28561, ____, _____
Given that:

13,169,2197,28561, ____, _____

Solution:

Step 1:   The starting value of the pattern is 13

Step 2:   The next value of the pattern = 13* 13 = 169

Step 3:   The third value can be written as  169 * 13 = 2197

Step 4:   The fourth value is 2197 * 13 = 28561

Now the left out value can be written as

28561 * 13 = 371293

371293* 13 = 4826809

So the correct pattern is  13,169,2197,28561,371293 ,4826809

The required solution is 13,169,2197,28561,371293 ,4826809

Substitute A with B

Introduction to substitute a with b:

Substitution is one of the methods used in math to find the value of unknown variable. When this process takes place some of the arithmetic operations like addition, subtraction, multiplication, division can be done. Finally the result will be a finite. Substituting a with b means the equation to find the value of a should have a variable b. Is this topic Linear Equation Solver hard for you? Watch out for my coming posts.

Example Problem – Substitute a with B.

Example 1 – Substitute a with b.

Find the value of a from the expressions 2a+10b = 18 and 5b = 20.

Solution:

The given expressions are 2a+10b = 18 and 5b = 20

To find the value of a it needs to calculate the value for b

5b = 20

Divide by 5 on both sides

b = 4.

Now plug in the value of b in the equation 2a+10b=18

2a+10(4) = 18

2a+40 = 18

Subtract 40 on both sides

2a+40 -40= 18-40

2a = -22

Divide by 2 on both sides

a = -11

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Two more Example Problem – Substitute a with B.

Example 2 – Substitute a with b:

Find the value of a from the expression 10a+4 = 14a+8b and 3b+4 = 10

Solution:

The given expressions are 10a+4 = 14a+8b and 3b+4 = 10

To find the value of a it needs to find the value of b from the equation 3b+4 = 10

Subtract 4 on both sides

3b+4 -4= 10-4

3b = 6

Divide by 3 on both sides

b = 2

Now plug in b = 2 in the equation 10a+4 = 14a+8b

10a+4 = 14a+8(2)

10a+4 = 14a +16

Subtract 14a on both sides

10a+4-14a = 14a +16-14a

-4a+4=16

Subtract 4 on both sides

-4a =12

Divide by -4 on both sides

-a = 3

a = -3.

Example 3 – Substitute a with b:

Find the value of a from the expressions 5a+10 = 12b and b = 17

Solution:

The given expressions are 5a+10 = 12b and b = 17

Now plug in b= 17 in the equation 5a+10 = 12b to find the value of a

5a+10 = 12(17)

5a+10 = 204

Subtract 10 on both sides

5a = 194

Divide by 5 on both sides

a = 38.8

Box and Whisker Graph Example

Introduction to box and whisker graph example:
A box and whisker graph is used to show a set of data so that you can simply see where most of the numbers are. Box plots are helpful in Quality Analysis for interpreting the division of data since it can easily show whether the data is distorted and if there are abnormal observations (outliers) in the dataset. Box plots are also very useful when huge numbers of observations are involved and when two or more datasets are being compared. The box and whisker plot  is a graphical representation of key values from abstract figures.

Methodology for Box and Whisker Plot:
In Box-and-whisker plot, the interior box represents the values from the lesser to higher quartile. The core line represents the median. The straight line extends from the minimum to the maximum value. Box plots show differences between datas without building any assumptions of the underlying statistical allocation. The spacing’s between the different parts of the box help indicate the degree of dispersion (spread) and skewness in the data, and identify outliers. Box plots can be drawn either parallel or perpendicularly. I have recently faced lot of problem while learning Median Statistics, But thank to online resources of math which helped me to learn myself easily on net.

Steps to Solve Box and Whisker Plot:

Step 1:The first step involves arranging the given data in an ascending order

Step 2:The second step involves find the median (Q1) for the entire data.

Step 3:After finding the median ,find the upper quartile and lower quartiles.

Step 4:Find the median for lower quartile (Q2) and upper quartile(Q3).

Step 5:Fiind the Minimum value and Maximum value.

Step 6:Plot these values for Q1,Q2,Q3 in the graph.

Step 7:Draw a box which joins all the median values.

Step 8:Draw a straight line from Minimum to maximum range.

Example:

Draw a  Box and whisker plot for the given data:

7,8,5,3,12,14,6,8,4,15,10

Step1: The first step involves arranging the given data in an ascending order

3,4,5,6,7,8,8,10,12,14,15

Step 2:The second step is to find the median (Q1) for the entire data.

Median Q1=8

Step 3:After finding the median ,find the upper quartile and lower quartiles

Lower quartile =3,4,5,6,7

Upper quartile = 8,10,12,14,15

Step 4:Find the median for lower quartile (Q2) and upper quartile(Q3).

Q1=5

Q­2=12

Step 5: Draw the graph and mark the positions of Q1,Q2 and Q3.

Step 6: Draw the box which joins Q1,Q2, and Q3.

Step 7: And draw the line which joins Minimum and Maximum value.

Probability Distribution Coin Toss

Introduction to probability distribution coin toss:

In probability, a probability distribution used to identify with in a particular interval falling value of a probability. The probability distribution describes the range of the possible value of a random variable. The probability distribution is entirely described by the cumulative distribution function. In this article we shall discuss probability distribution coin toss example problem.

Probability Distribution Coin Toss Example Problem

Example: A coin is tossed for 65times. Find the expectation value of tail with the help of using the binomial distribution.

Solution:

Consider 65 times the coin is tossed, so n = 65

When toss a coin, we get probability of tail is p = 1/2

Probability of getting tail is represent by q

q = 1 – p                        [p + q = 1]

q = 1 – 1/2

q = 1/2

Expectation value:

μ = E[x] = np

= 65(1/2 )

= (65/2)

μ = E[x] = 32.5

Answer:

Expectation value = μ = E[x] = 32.5

Example 2: An unbiased die is rolling for 45 times. By using the binomial distribution find the expectation value of getting four?

Solution:

Let p be probability of getting success

Let q = 1 – p probability of getting failure

Let n be number of times the dice is thrown

An unbiased dice is rolling for 45 times, the value of n = 43

When we throe a unbiased dice, probability of getting four is p = 1/6

Probability of not getting four is represent by q

q = 1 – p

= 1 – 1/6

= 6/6– 1/6

= 6-1/6

q = 5/6

Expectation value:

μ = E(x) = np

= 45 (1/6)

= 45/6

= 7.5

μ = E(x) = 7.5

Answer:

Expectation value = μ = E[x] = 7.5

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Probability Distribution Coin Toss Practice Problem

Problem:

A coin is tossed for 42 times. Find the expectation value of tail with the help of using the binomial distribution.

Answer:

Expectation value = 21

Problem:

An unbiased die is rolling for 35 times. By using the binomial distribution find the expectation value of getting four?

Answer:

Expectation value = 5.8

Mixed Probability Distribution

Introduction to mixed probability distribution:

In this article we shall discuss the mixed probability distributions. Probability plays the main part in the mathematics and in all new fields. Probability is used to discover number of incidence of an event out of total outcomes. The probability of certain two random variables X and Y define on the equivalent probability space, the mixed distribution for X and Y defines the probability of events distinct in expressions of together X and Y.

Basic for Mixed Probability Distribution:

We found with a random test among probability measure P on a primary sample space. Now, we will confer two "mixed" cases for the probability distribution of a random variable in this case where the distribution is partly discrete and partly continuous and the case where the variable has both discrete coordinates and continuous coordinates.

Mixed Probability Distributions Consist of Following Cases:

Cumulative distribution case:

The cumulative distribution function for a couple of random variables is distinct in terms of their mixed probability distribution,

`F(x,y)=P(Xlt=x,Ylt=y)` .

Discrete distribution case:

For discrete random variables, the mixed probability distribution is,

`P(X=x and Y=y)=P(Y=y|X=x).P(X=x)` .

`=P(X=x|Y=y).P(Y=y)` .

Since these are probabilities, we have

`sum_(x)` `sum_(y)`` P(X=x and Y=y)=1` .

Continuous distribution case:

Likewise for continuous random variables, the mixed probability distribution function can be denoted as fX,Y(x, y) and this is,

`f_(X.Y)(x,y)=f_(Y|X)(y|x)f_(X)(x)=f_(X|Y)(x|y)f_(Y)(y)` .

Anywhere fY|X(y|x) and fX|Y(x|y) provide the conditional distribution of Y known X = x and of X known Y = y in that order, and fX(x) and fY(y) give the secondary distribution for X and Y correspondingly. Is this topic math 4th grade word problems hard for you? Watch out for my coming posts.

Again, since these are probability distributions, one has

`int_(x)int_(y)f_(X,Y)(x,y)dydx=1` .

Mixed probability distribution case:

Most of time X is continuous but Y is discrete. Intended for case, in a logistic regression, one might wish to predict the probability of a binary outcome Y conditional on the value of a continuously-distributed X. On the other hand, a "mixed probability distribution" can be distinct in together of two ways.

`f_(X,Y)(x,y)=f_(X|Y)(x|y)P(Y=y)` .

`=P(Y=y|X=x)f_(X)(x)` .

Properly, fX,Y(x, y) is the probability mass function of (X, Y) with value to the product measure on the particular supports of X and Y. each of these two decompositions can then be used to pick up the varied cumulative distribution function.

`F_(X,Y)(x,y)=sum_(tlt=y)int_(s=-oo)^x f_(X,Y)(s,t)ds` .

Grouping of arbitrary statistics of discrete and continuous random variables.