Homogeneous Function
Homogeneous function is a function with multiplicative scaling behaving. The function f(x, y), if it can be expressed by writing x = kx, and y = ky to form a new function f(kx, ky) = knf(x, y) such that the constant k can be taken as the nth power of the exponent, is called a homogeneous function.
Let us learn more about homogeneous function, and the differential equation of a homogeneous function, with examples, FAQs.
1. | What Is A Homogeneous Function? |
2. | Homogeneous Differential Equation from Homogeneous Function |
3. | Examples on Homogeneous Function |
4. | Practice Questions |
5. | FAQs on Homogeneous Function |
What Is A Homogeneous Function?
Homogeneous function is a function with multiplicative scaling behaving. Here if each variable in the equation is multiplied with a constant, then the entire function is also multiplied with a exponent of the constant value. Let us consider a function f(x, y), and if each variable is multiplied with a constant K, then the entire function expression is also multiplied with the nth power of the constant k.
f(kx, ky) = knf(x, y)
Here the exponent n is called the degree of homogeneity.
f(x, y) = x2 - 2xy, we have f(kx, ky) = (kx)2- 2(kx)(ky) = k2x2 - 2k2xy = k2(x2 - 2xy)
f(x, y) = 3x + 4y, we have f(kx, ky) = 3(kx) + 4(ky) = 3kx + 4ky = k(3x + 4y)
f(x, y) = Tan(x/y), we have f(x, y) = Tan(kx/ky) = Tan(x/y) = k0Tan(x/y)
Here in the above functional equations we observe that the constant k can be taken common and hence the above functions are homogeneous functions of degree 2, 1, 0 respectively.
f(x, y) = xLogy + 5xy, and we have f(kx, ky) = kx.Logky + 5(kx)(ky) = kx.Logky + 5k2xy
f(x, y) = 2Sinx + Cosx , and we have f(kx, ky) = 2Sinkx + Coskx
In the above two functions we cannot easily take the constant common, and hence the above two functions cannot be considered as homogeneous functions. Function containing special expressions involving logarithms or trigonometric ratios are not homogeneous functions.
Homogeneous Differential Equation From Homogeneous Function
The differential equation formed using a homogeneous function is a homogeneous differential equation. The differential equation of the form dy/dx = f(x, y) is a homogeneous differential equation if the function f(x, y) is a homogeneous function. Further, let us check an alternate definition of a homogeneous function before we progress to solve a homogeneous differential equation.
A function f(x, y) which can be expressed in the form f(x, y) = xn.g(y/x), or yn.h(x/y) is a homogeneous function of degree n. For solving a homogeneous differential equation of the form dy/dx = f(x, y) = g(y/x) we need to substitute y = vx, and differentiate this expression y = vx with respect to x. Here we obtain dy/dx = v + x.dv/dx. Let us substitute this value of dy/dx in the differential equation dy/dx =f(x, y) = g(y/x).
v + x.dv/dx = g(v)
x.dv/dx = g(v) - v
dv/(g(v) - v) = dx/x
Here we have successfully separated the variables. This can be further integrated to find the solution of this differential equation.
\(\int \dfrac{dv}{g(v) - v} = \int \dfrac{1}{x}.dx + C\)
This equation gives the general solution of the differential equation if we replace back v = y/x. Also if the homogeneous differential equation is of the form dx/dy = f(x, y), and if f(x, y) is a homogeneous differential equation, then we substitute x = vy, to find the general solution of the differential equation dx/dy = f(x, y) = h(x/y).
Related Topics
The following topics would help in a better understanding of homogeneous function.
Examples on Homogeneous Function
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Example 1: Find if the function f(x, y) = x3+ 2x2y - 3xy2 + y3 is a homogeneous function.
Solution:
The given expression is f(x, y) = x3+ 2x2y - 3xy2 + y3
Let us substitute x = kx, and y = ky in the above expression.
f(kx, ky) = (kx)3+ 2(kx)2(ky) - 3(kx)(ky)2 + (ky)3
f(kx, ky) = k3x3+ 2k3x2y - 3k3xy2 + k3y3
f(kx, ky) = k3(x3+ 2x2y - 3xy2 + y3)
f(kx, ky) = k3f(x, y)
Therefore, the above function is a homogeneous function.
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Example 2: Find the solution of the homogeneous differential equation dx(x2 + y2) = dy.2xy
Solution:
The given differential equation is dx(x2 + y2) = dy.2xy.
\(\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{2xy}\)
\(\dfrac{dy}{dx} = \dfrac{1+ \frac{y^2}{x^2}}{2\frac{y}{x}}\)
Let us replace y = vx in the above expression.
\(\dfrac{dy}{dx} = v + x \dfrac{dv}{dx}\)
\(v + x \dfrac{dv}{dx} = \dfrac{1 + v^2}{2v}\)
\(x\dfrac{dv}{dx} = \dfrac{1 - v^2}{2v}\)
\(\dfrac{2v}{1 - v^2}.dv = \dfrac{dx}{x}\)
\(\dfrac{2v}{v^2 - 1}.dv = -\dfrac{dx}{x}\)
\(\int \dfrac{2v}{v^2 - 1}.dv = -\int \dfrac{dx}{x}\)
Log(v2 - 1) =- Logx + Log C
Log(v2 - 1)x = LogC
Log(y2/x2 - 1)x= LogC
Log(y2 - x2)/x = LogC
y2 - x2 = xC
x2 - y2 = xC.
Therefore the solution of the differential equation is x2 - y2 = xC.
FAQs on Homogeneous Function
What Is A Homogeneous Function?
The homogeneous function is a function with multiplicative scaling behaving. Here if each variable in the equation is multiplied with a constant, then the entire function is also multiplied with an exponent of the constant value. For a function f(x, y), and if each variable is multiplied with a constant K, then the entire function expression is also multiplied with the nth power of the constant k.
f(kx, ky) = knf(x, y)
How Do You Know If A Function Is A Homogeneous Function?
The function f(x, y) if it can be expressed in the form f(kx, ky) = knf(x, y) such that the constant k can be taken common with the nth power of the constant is a homogeneous function equation.
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