subject: Solving Homogeneous Equation [print this page] 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
