Solving Simple Differential Equation

Solving simple differential equations involve the process of differentiating the algebraic function with respect to the input function. The algebraic function which is differentiable is known as differential equations. The differential equation comes under calculus category whereas to find the rate of change of the given function with respect to the input function. The following are simple example differential equations for solving.


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Simple Differential Equations Examples for Solving:

The following are the example problems with simple differential equations for solving.

Example 1:

Solve the simple differential equation.

f(k) = k2 – 4k + 8

Solution:

The given equation is

f(k) = k 2 – 4k + 8

The first derivative f ' for the algebraic function is

f '(k) = 2 k  – 4

Example 2:

Solve the simple differential equation.

f(k) = k 3 – 5 k 2  + 11k

Solution:

The given function is

f(k) = k 3 – 5 k 2  + 11k

The first derivative f ' for the algebraic function is

f '(k) = 3k 2 – 5(2 k  ) + 11

f '(k) = 3k 2 – 10 k + 11

Example 3:

Solve the simple differential equation.

f(k) = k4 – 3k 3 – 4k 2  + k

Solution:

The given function is

f(k) = k4 – 3k 3 – 4 k 2  + k

The first derivative f ' for the algebraic function is

f '(k) = 4 k 3 – 3(3k 2 ) – 4( 2 k  ) + 1

f '(k) = 4 k 3 – 9k 2  – 8 k  + 1

My forthcoming post is on Example of Hypothesis Testing and Dividing Fraction will give you more understanding about Algebra.

Example 4:

Solve the simple differential equation.

f(k) = k 5 – 6 k 3  + 11

Solution:

The given function is

f(k) = k 5 – 6 k 3  + 10

The first derivative f ' for the algebraic function is

f '(k) = 5k 4 – 6(3 k 2 )

f '(k) = 5k 4 – 18 k 2
Simple Differential Practice Equations for Solving:

1) Solve the simple differential equation.

f(k) = k 3 – 6 k 2  + 11k

Answer: f '(k) = 3k 2 – 12 k

2) Solve the simple differential equation.

f(k) = k 2 – 6 k   + 11

Answer: f '(k) = 2k – 6

About Metric Volume Units

The quantities used to find area, length, width capacities and volume of things etc are called measures. Many countries have their own system of measures. But Metric System of measures is very simple and easy to calculate. The area is measured in square unit. In metric system the volume is measured in cubic units.
Example Problems - Metric Volume Units:

The triangular prism has width 6 cm, height 9 cm and length 11 cm. find the volume of triangular prism.

Solution:

Given:

Width (w) = 6 cm

Height (h) = 9 cm

Length (l) = 11 cm

Formula:

Volume of triangular prism (V) = `1/2` (l x w x h) cubic units

= `1/2` (11 x 6 x 9)

= `1/2` (594)

= 297

Volume of triangular prism (V) = 297 cm3

2. figure out the volume trapezoidal prism whose length 11 cm, height 8cm, length of parallel sides a=7 cm and b=4cm.

Solution:

Given:

Length (l) = 11cm

Height (h) = 8cm

Parallel sides a=7cm and b=4cm

Formula:

Volume of trapezoidal prism = l x area of the base cubic units

Area of the base:

Area of the base = `1/2` x (a + b) x h

= `1/2` x (7 + 4) x 8

=`1/2` x 11 x 8

= 44 cm2

Volume of trapezoidal prism = 11x 44

= 484

Volume of trapezoidal prism = 484 cm3

3. The cylinder has the radius r = 10 feet, h=23 feet. Find the volume of cylinder.

Solution:

Given:

r=10 cm

h=23 cm

Formula:

The volume of the cylinder = π x r2 x h cubic unit

=3.14 x (10)2 x 23

The volume of the cylinder = 7222 ft3.
Example Problems - Metric Volume Units:

Cone:

4. The cone has the radius = 10 feet and height = 23 feet. Find the volume of the cone.

Solution:

Given:

Radius (r) = 10 feet

Height (h) = 23 feet

Formula:

The volume of the cone =`1/3` x π x r2  x  h

= `1/3` x 3.14 x (10)2 x 23

The volume of the cone = 2407.33 ft3

5. What is the volume rectangular prism with length 8 cm width 5 cm and height 6 cm?

Solution:

Given:

Length =8 cm

Width = 5 cm

Height = 6 cm

Formula:

Volume of rectangular solid (v) = l x w x h

= 8 x 5 x 6

= 240

Volume of rectangular solid (v) = 240 cm3

radius calculation formula

Pi is also one of the important concepts in algebra. The value of pi is 22/7 or 3.14. Using these values we can easily find the answer for complex problems in algebra. We have different kinds of problems using pi in algebra. Commonly the pi formulas are used to find area or volume of the specific shape in algebra. Here we are going to learn some problems for calculate radius.
Radius Calculation Formulas:

Perimeter of a circle = 2 * pi * radius.
Area of circle = Pi * radius 2.
Area of ellipse = pi *radius1 *radius 2.
Surface area of sphere = 4*pi* radius 2.
Surface area of cylinder = 2 * pi *radius *height.
Surface area of cone: pi* radius* side.
Surface area of torus = pi2 * (radius2 2 –radius12).
Volume of sphere = `(4)/(3)` *pi*radius 3.
Volume of ellipsoid = `(4)/(3)` *pi*radius 1*radius 2*radius3.
Volume of cylinder =pi*radius*height.
Volume of cone = `(1)/(3)` * pi* radius 2 *height.
Volume of torus = `(1)/(4)` * pi2 *(radius1 + radius2) *(radius1-radius2)2
Volume of hemisphere = `(2)/(3)` pi r3.

These are the formulas to calculate radius.
Example Problems of Radius Calculation Formula:

Example 1:

Perimeter of a circle is 40 m .calculate the radius of the circle?

Solution:

Step 1: Perimeter of a circle = 2 * pi * radius.

Step 2: We know that perimeter = 40 m.

Step 3: Plug the perimeter value in to the formula.

Step 4: 40 = 2 * 3.14 * radius.

Step 5: 40 = 6.28 *radius. (Divide using 6.28 on both the sides)

Step 6: Therefore, radius = 6.4 m.

Example 2:

Surface area of cone is 120 cm^2 and side = 7cm.Calculate the radius of the cone?

Solution:

Step 1: Surface area of cone: pi* radius* side.

Step 2: We know that surface area and side.

Step 3: Plug the surface area and side value in to the formula.

Step 4: Therefore, surface area of cone is 120 = 3.14 * radius *7.

Step 5: So, 120 = 21.98 *radius. (Divide using 21.98 on both the sides)

Step 5: Therefore radius = 5.5 cm.

These are the example problems of calculate radius using formula.

Practice problems of calculate radius using formula.

1)   Volume of cylinder = 160 m and height = 9 cm .calculate the radius of the cylinder?

2)   Suppose the area of circle is 210 m. calculate the radius of the circle?

Answer key

1)   Radius = 5.7 m

2)   Radius = 8.17 m

Dividing Decimal Word Problems

Decimal is one of the number systems which have the base value 10. Division is one of the basic arithmetic operations which is the inverse of multiplication operation. Dividing decimal word problems occurs in such circumstances where we are trying to find out how many times a decimal number go into another. In this article we will see few examples of dividing decimal word problems.

Dividing Decimal Word Problems Examples:

Example 1:

If 2.5 kg of apples cost `$` 15.50 what is the price of one kilogram of apple?

Solution:

Here to find the cost of 1kg of apple we need to divide 15.50 by 2.5

Division of 15.50 by 2.5

By dividing `15.50/2.5` = 6.2

Hence the price of 1 kg apple = `$` 6.2

Example 2:

Tom wants to buy chocolates. The cost of one chocolate is `$` 0.5. The total amount tom has is `$` 20.25. How many chocolates he can buy?

Solution:

Total amount Tom has =` $` 20.25

Cost of one chocolate = `$` 0.5

Number of chocolate can buy =` 20.25 / 0.5`

Division of 20.25 by 0.5

= 40.5

Hence Tom can buy 40 chocolates for `$` 20.25

Example 3:

Each bag of cherries weighs 2.25 pounds. How many bags of cherries would be needed to make 92.25 pounds?

Solution:

Weight of 1 bag of cherries = 2.25 pounds

Total pounds = 92.25

To find number of bags of cherries,

Divide 92.25 by 2.25

Division of 92.25 by 2.5

By dividing` 92.25 /2.25` = 41

Hence 41 bags of cherries needed to make 92.25 pounds.


Dividing Decimal Word Problems Examples Continued:

Example 4:

There are 15 students in a class.  Together they weigh 1925.90 pounds.  Find the average weight?

Solution:

Total Number of students = 15

Total weight = 1925.90 pounds

Average weight = `1925.90/15`

Division of 1925.90 by15

=128.39pounds

Example 5:

There are 2.54 centimeters in one inch.  How many inches are there in 75.24 centimeters?

Solution:

1 inch = 2.54cm

75.24cm= `75.24/2.54`

Division of 75.24 by 2.54

= 29.62inches

Hence there are 29.62inches in 75.24cm

Example 6:

The scores of three persons in diving competition are 3.4, 5.75, and 6.61.  What is the average score?

Solution:

Average score = Total score/ number of persons

Total score =   3.4+5.75+6.61=15.76

Average score = `15.76/3`

Division of 15.76 by3

= 5.25

Two Equations in Standard Form

Linear equation is an algebraic equation which has constants and variables together. Linear equation has 1 or more variables. Linear equation has more forms. Standard form is one of the form of linear equation.

Standard form of linear equation is Ax + By = C.

Here A, B and C are constants. X and y are variables. A and B are not zero.

We can solve standard form of equation using substitution or elimination method. But here we use two standard form equations. Let us see how to solve.

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Standard Form of Two Equations – Substitution Method:

Problem:

Solve these two standard forms of equation using substitution method.

2x + 4y – 12 = 0

X – 2y – 4 = 0

Solution:

The given standard form of two equations are,

2x + 4y = 12  (1)

X – 2y = 4  (2)

From (2), we rewrite the equation as

X = 4 + 2y  (3)

Substitute equation (3) into (1) to find the variable y.

2(4+2y) + 4y = 12

Apply the distributive property, we get

8 + 4y + 4y = 12

Combine like terms,

8+ 8y = 12

Subtract 8 from each side.

8 – 8 + 8y = 12 – 8

8y = 4

Divide by 8 each side.

`(8y)/8` = `4/8`

Y = `1/2`

Substitute y = `1/2` into equation (2)

X – 2y = 4

X – 2(`1/2` ) = 4

X – 1 = 4

Add 1 to each side.

X – 1 + 1 = 4 + 1

X = 5.

Therefore, the solutions are 5 and 1/2.

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Standard Form of Two Equations – Elimination Method:

Problem :

Use elimination method to determine the solutions of the following the systems of equations.

x + y – 16 = 0 and 4x – 2y – 4 = 0

Solution:

The given standard forms of two equations are

x + y = 16  (1)

4x – 2y = 4  (2)

Step 1:

Multiply the equation (1) by 2 and Equation (2) by 1 to get the coefficients of variable y same. So the equations are,

2x + 2y = 32

4x – 2y = 4

Step 2:

Add the two equations for eliminating y variable.

2x + 2y + 4x – 2y = 32 + 4

2x + 4x + 2y – 2y = 36

6x = 36

Divide by 6 both sides.

x = 6

Step 3:

Substitute the x value into the equation (1) to get value of y variable.

x + y = 16

6 + y = 16

Subtract 6 from each side.

y = 10.

The solutions are x = 6 and y = 10.

Rational Zeros of a Function

Normally zeros of a function mean when we plug the values for the variables the functions values tends to be zero.  Let us consider if we are having a function with variable x and we have a set of solution p(x) if we plug the solution for the given variables present in the function we will get the function f(x) = 0. To find the rational zeros we have to use the rational zeros theorem.

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Rational Zeros of a Function – Examples:

The rational theorem states that if we are having the polynomial p(x) wit the integer coefficients and we are having the zeros of the polynomial `p / q` then we can say `p(p / q) = 0` . Here p is nothing but the constant term of the polynomial and the q I nothing but the leading coefficient of the polynomial p(x). We will see some examples for finding the rational zeros of a function.

Example 1 for rational zeros of a function:

Find the rational zeros of the following function.

2x2 + 12x + 10

Solution:

The given function is

2x2 + 12x + 10

First we have to find the rots of the constant term. ±1, ±2, ±5, ±10

Now the leading co – efficient of the constant term is 2. So we have to divide by 2.

±`p / q` = ±`1/ 2` , ±`2/ 2` , ±`5/ 2` , ±`10/ 2`

= ±`1/2` , ±1, ±`5/2` , ±5

Now we have to use the synthetic division method to find the rational zeros.

1 / 2 | 2           12        10

|               1         13/2

|_____________________

2           13        33/2      = not a zero

-1 / 2 | 2           12        10

|               -1         -11/2

|_____________________

2           11        9/2      = not a zero



1 | 2           12        10

|               2         24

|_____________________

2           24        34      = not a zero

-1 | 2           12        10

|               -2        -10

|_____________________

2           10        0      = is a zero

5/2 | 2           12        10

|               5        85/2

|_____________________

2           17        105/2      = not a zero

- 5/2 | 2           12        10

|               -5        -35/2

|_____________________

2           7        15/2      = not a zero

5 | 2           12        10

|              10        110

|_____________________

2           22        120      = not a zero

-5 | 2           12        10

|              -10      -10

|_____________________

2           2        0      = is a zero

So from the above the rational zeros of the functions are p(x) is -1 and -5

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Rational Zeros of a Function – more Examples:

Example 2 for rational zeros of a function:

Find the rational zeros of the following function.

x2 + 4x + 3

Solution:

The given function is

x2 + 4x + 3

First we have to find the rots of the constant term. ±1, ±3

Now the leading co – efficient of the constant term is 1. So we have to divide by 1.

±`p / q` = ±`1/ 1` , ±`3 / 1`

= ±1, ±3

Now we have to use the synthetic division method to find the rational zeros.

1 | 1           4        3

|               1       5

|_____________________

1           5       8      = not a zero

-1| 1           4        3

|              -1      -3

|_____________________

1            3        0      = is a zero

3 | 1           4        3

|              3        21

|_____________________

1          7       24      = not a zero

-3| 1           4        3

|             -3      -3

|_____________________

1            1        0      = is a zero

So from the above the rational zeros of the functions are p(x) is -1 and -3

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Interval of Convergence for Taylor Series

The interval of convergence for the given series is the set of all values such that the series converges if the values are within the interval and diverges if the value exceeds the interval.
The interval of convergence series must have the interval a - R < x < a + R since at this interval power series will converge.

In this article, we are going to see few example and practice problems of Taylor series to find interval of convergence which help you to learn interval of convergence.
Example Problems to Find the Interval of Convergence for Taylor Series:

Example problem 1:

Solve and determine the interval of convergence for Taylor seriessum_(n = 0)^oo(x^n)/(n!) .

Solution:

Step 1: Given series

sum_(n = 0)^oo (x^n)/(n!)  .

Step 2: Find L using the ratio test

L =  lim_(n->oo) | ((x^(n+1))/((n+1)!))/((x^n)/(n!)) |

= lim_(n->oo) |   (xn!)/((n + 1)(n!))  |

= lim_(n->oo) |   x/(n + 1)  |

= 0

So, this series converge for all value of x. Therefore, the interval of convergence is (-∞, ∞).

Step 3: Solution

Hence, the interval of convergence for the given Taylor series is (-∞, ∞).

Example problem 2:

Solve and determine the interval of convergence for Taylor series sum_(n = 0)^oo(-1)n (x^(2n + 1))/((2n + 1)!) .

Solution:

Step 1: Given series

sum_(n = 0)^oo(-1)n (x^(2n + 1))/((2n + 1)!)  .

Step 2: Find L using the ratio test

L =  lim_(n->oo) | (-1)^(n + 1)(x^(2(n + 1)+1))/((2n + 1)!) |

= lim_(n->oo) |  (-1)^(n + 1)(x^(2n + 3))/((2n + 1)!)   |

= 0

So, this series converge for all value of x. Therefore, the interval of convergence is (-∞, ∞).


Step 3: Solution

Hence, the interval of convergence for the given Taylor series is (-∞, ∞).

Practice Problems to Find the Interval of Convergence for Taylor Series:

1) Determine the interval of convergence for Taylor series sum_(n = 0)^oo(-1)n (x^(2n))/(2n!)  .

2) Determine the interval of convergence for Taylor series sum_(n = 0)^ooxn.

Solutions:

1) The interval of convergence for the given Taylor series is (-∞, ∞).

2) The interval of convergence for the given Taylor series is |x| < 1.