The homogeneous equation can be transferred into the changes of variables.

A differential equation of first order and first degree is said to be homogeneous if it can be put in the form

'(dy)/(dx)' = f(y/x) or

'(dx)/(dy)' =f1(x, y)/f2(x, y)

Working rule for solving homogeneous equation : By definition the given equation can be put in the form

'(dy)/(dx)' = f (y/x) (1)

To solve y = x (2)

Differentiating (2) with respect to x gives

'(dy)/(dx)' = + x(d/dx) (3)

Using (2) and (3) in (1) we have

+ x'(dy)/(dx)' = f() or x'(dy)/(dx)' = f()

Separating the variables x and we have

(dx/)x =(d/f() )

= log x + c

= d/(f() )

where c is an arbitrary constant. After integration, replace by y/x .

Solving Homogeneous Equation-problem

Solve : =(y/x) + tan(y/x)

Solution : Put y = vx

L.H.S. = + x'(dy)/(dx)' ;

R.H.S. = v + tan v

Therefore + x'(dv)/(dx)'

= + tan or

=dx/x

= (cos /sin) dv

Integrating, we have logx = log sin + log c x = c sin

i.e., x = c sin (y/x).

Solving Homogeneous Equation-homogeneous Problem

Solve : (2 xy x) dy + ydx = 0

Solution : The given equation is '(dy)/(dx)' =(-y/(2 xy x)

Put y = vx

L.H.S. = v + x'(dy)/(dx)'

R.H.S. = v/2( v 1)

= v/(1 2 v)

Therefore v + x'(dv)/(dx)'

=v/(1 2 v)

x'(dv)/(dx)'

=2 v v/(1 2 v)

=(1 2 v)/vv)dy

= 2 dx/x

=(v3/2 2.1/v) dv

= 2dx/x

= 2v1/2 2 log v = 2 log x + 2 log c

= v1/2 = 2log (v x c)

= xy = log(cy)

= cy = e x/y or yex/y

= c

This problem can also be done easily by taking x = vy.

Pre calculus, (or Algebra 3 in some areas) an advanced form of secondary school algebra, is a foundational mathematical discipline. It is also called Introduction to Analysis. In many schools, pre calculus is actually two separate courses: Algebra and Trigonometry. Pre calculus does not prepare students for calculus as pre-algebra prepares students for Algebra. While pre-algebra teaches students many fundamental algebras topics, pre calculus does not involve calculus, but explores topics that will be applied in calculus.

Example for Solving Pre-calculus Equation:

SIMULTANEOUS EQUATIONS:

Problem:

Solving the equation : x+ y = 3, y + z = 5, z + x = 2.

Solution:

Let the equations be identified as

x + y = 3 (1)

y + z = 5 (2)

z + x = 2 (3)

Adding all the three equations, we get

2x + 2y + 2z = 3 + (5) + 2 or 2(x + y + z) = 0 or x + y + z = 0 (4)

Substituting y + z = 5 in equation (4) we get x + (5) = 0 x = 5.

Substituting z + x = 2 in equation (4) we get y + 2 = 0 y = 2.

Substituting x = 5 in equation (3) we get z + 5 = 2 or z = 25 = 3

The solution is x = 5, y = 2, z = 3.

Addition of Polynomial equation:

We add two polynomials by adding the coefficients of the like powers.

Problem :

Find the sum of 2x4 3x2 + 5x + 3 and 4x + 6x3 6x2 1.

Solution:

Using the associative and distributive properties of real numbers, we get

(2x4 3x2 + 5x + 3) + (6x3 6x2 + 4x 1) = 2x4 + 6x3 3x2 6x2 + 5x + 4x + 3 1

= 2x4 + 6x3 (3+6)x2 + (5+4)x + 2

= 2x4 + 6x3 9x2 + 9x + 2.

Quadratic equation:

Problem:

Solving given equation 15 2x x2= 0

Solution:

Writing in the standard form,

15 2x x2 = x2 2x + 15

= (1) (x2 + 2x 15).

Here, we find 15 = 5 3, 5 + (3) = 2

Hence, we get 15 2x x2 = (1) [(x+5) {x + (3)}]

= (1) (x +5)(x 3)

= (x + 5) ( 3 x)

X = -5 , -3

Solving Pre-calculus Equation-practice Problems:

1. Solving equations: x2- 8x + 16

Answer: x= 4 ,4

2. Find the product of 2x + 3y and x2 xy + y2.

Answer: 2x3 + x2y xy2 + 3y3.