Thursday, April 9

Question on Integrating Factor Method for Solving Differential Equation

Topic : Integrating Factor

Question : Explain Integrating Factor Method for Solving Differential Equation

Solution :
Suppose we have this equation to solve:
dy + y = x

dx
In this case we cannot separate the variables and introducing a new variable doesn't help either. So we need to use an entirely different method, known as Integrating Factor Method.
Here's the typical equation for which we can use this method:

dy + yP(x) = Q(x)
dx

Here the functions P(x) and Q(x) can be any functions of x, in the example above they were 1 and x respectively.
Step One: calculate the integral of the function P(x).
Step Two: the integrating factor, which we'll call IF, is defined as the exponential of this, i.e. it's defined by :

Integrating Factor = IF = e^(∫P(x) dx)
Step Three: the solution to the equation is given by:
y(x) = 1/IF ∫Q(x) IF dx
Now let us understand this using an example-
dy + y = x
dx

So P(x)=1 and Q(x)=x. The integrating factor is therefore ex, since the integral of 1 is x. (No need for a constant of integration, that's because if you do put one in it will cancel out later on anyway)
So, we can write down the solution y(x) of the equation using the "mysterious" formula given in Step Three:

y(x) = 1/IF ∫Q(x) IF dx
In this case that's:
y(x) = 1/e^x ∫xe^x dx
We can do that integral using integration by part. The result is:
y(x) = e^(-x) (xe^x - e^x + c)
which implies to
y(x) = x - 1 + ce^(-x)

Friday, April 3

Problem to Solve Simultaneous Equations by Substitution Method

Topic : Substitution Method

Problem : Solve using Substitution Method.
y - 2x = - 6
2y - x = 5

Solution :


Monday, March 30

Question to Find Factors of a Function

Topic : Functions

Question : Given f(x) = 2x³ - 5x² + 12x - 5 Find all zeros, if
p : ± 1, ± 5
q : ± 1, ± 2
p/q = ± 1/2 , ± 5/2

Solution :

Tuesday, March 24

Question on Linear Equation

Topic : Linear Equation

Question : Sue wants to buy a new printer that costs $189. She has $125 saved. She has a job that pays $8 an hour. How many hours must she work to earn enough to buy the printer?

Solution :
The cost of the printer is $189.
She has saved $125.
Also, she is paid $8 an hour to do a job.
We need to find the number of hours she works to buy the printer.
Let us assume “x” to be the number of hours she works to buy the printer.
8x + 125 =189
Now, subtract 125 on either side of the equation.
8x + 125 – 125 = 189 – 125
8x = 64
Divide by 8 on either side of the equation.
8x/8 = 64/8
x = 8
Therefore, the number of hours she works to buy the printer is 8.

Thursday, March 19

Solved variable Saparable form of Differential Equation

Topic : Variable Saparable Form

Question : Solve dy/dx = (x-1)/(y+2)

Solution :
dy/dx = (x-1)/(y+2)
(y+2) dy = (x-1) dx
∫(y+2) dy = ∫(x-1) dx
y²/2 + 2y = x²/2 - x + c1 (c1 is the constant)

Monday, March 16

Problem on Simple Differentiation

Topic : Simple Differentiation
Problem : Solve y = x² + 4e^x
Solution :
y = x² + 4e^x
dy/dx = d(x²)/dx + 4d(e^x)/dx
dy/dx = 2x + 4e^x

Wednesday, March 11

Word Problem on Sine Rule

Topic : Sine Rule

Question :
Lookout station B is located 7 miles due east of station A. The bearing of a fire from A is S 8degree50’ W and the bearing from B is S 28degree20’ W. Determine the distance from the fire to B(to the nearest tenth of a mile)

Solution :


Figure is as shown in the figure according the question



From Sine Rule

a / Sin A = b / Sin B = c / Sin C

a = c Sin A / Sin C
a = 7 * Sin (90º + 8º 50’) / Sin (90º - 28º 20’)
a = 7 * Sin (98º 50’) / Sin (19º 30’)

a = 20.72miles