3 Homogeneous Function
The function \(f(x,y)\); if it can be expressed by writing \(x = kx\) and \(y = ky\) to form a function \(f(kx,ky)= k^{n} f(x,y)\), such that constant \(k\) can be taken as the \(n^{th}\) power of exponent , it is called as homogeneous function of degree \(n\). Here \(n\) is called as the degree of homogeneity.
Example 3.1: Verify \(f(x,y)\) is a homogeneous function.
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\(f(x,y) = x^{2} - 2xy\)
\(f(kx,ky) = (kx)^{2}- 2(kx)(ky)\)
= \(k^{2} x^{2} - 2k^{2}xy\)
= \(k^{2}(x^{2}- 2xy)\)
= \(k^{2} f(x,y)\)
Homogeneous function of degree 2.
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\(f(x,y)= 3x+4y\)
\(f(kx,ky)= 3(kx)+4(ky)\)
\(= 3kx+4ky\)
\(= k(3x+4y)\)
\(= k f(x,y)\)Homogeneous function of degree 1.
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\(f(x,y)= tan(x/y)\)
\(f(kx,ky)= tan(kx/ky)\)
\(= k^{0} tan(x/y)\)
\(= k^{0} f(x,y)\)Homogeneous function of degree 0.
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\(f(x,y)= x log y+ 5xy\)
\(f(kx,ky)=kx. log ky+ 5 kx. ky\)
\(= kx. log y+ 5k^{2} xy\)Not a Homogeneous function
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\(f(x,y)= 2 sin x+ cos y\)
\(f(kx,ky)=2 sin kx+ cos ky\)
Not a Homogeneous function
Problem 3.1: Verify wether the function below is homogenous
\(f(x,y)= x^{3}+ 2x^{2}y- 3xy^{2} + y^{3}\)
3.1 Homogeneous differential equation
\(\frac{dy}{dx} = f(x,y)\) is a homogeneous differential equation if \(f(x,y)\) is a homogeneous function.
3.2 Alternative definition of homogeneous function
A function \(f(x,y)\) is homogeneous function of degree \(n\) if it can be expressed in the form
\(f(x,y)= x^{n} g(y/x)\)
\(f(x,y)= y^{n} h(x/y)\)
3.2.1 Solving a homogeneous differential equation
\(\frac{dy}{dx} = f(x,y)= g(y/x)\) (1)
substitute \(y= vx\).
\(\frac{dy}{dx} = v+ x.\frac{dv}{dx}\) (2)
substitute (2) in (1)
\(v+ x.\frac{dv}{dx} = g(v)\)
\(x.\frac{dv}{dx}= g(v)- v\)
\(\frac{dv}{g(v)-v} = \frac{dx}{x}\)
this can be further integrated to find the solution of differential equation.
\(\int\frac{dv}{g(v)-v} = -\int\frac{1}{x} dx+ c\)
general solution is obtained by replacing back \(v= y/x\)
3.2.1.1 QUESTION 1.
Find the solution of homogeneous differential equation
\(dx (x^{2} + y^{2})= dy. 2xy\)
Solution
\(\frac{dy}{dx}= \frac{1+\frac{y^2}{x^2}}{2\frac{y}{x}}\) (I)
let \(y= vx\)
\(\frac{dy}{dx}= v+ x. \frac{dv}{dx}\) (II)
(II) in (I)
\(v + x\frac{dv}{dx} = \frac{1+\frac{v^2x^2}{x^2}}{2\frac{vx}{x}}\)
\(= \frac{1+v^2}{2v}\)
\(x\frac{dv}{dx} = \frac{1+v^2}{2v}-v\)
\(= \frac{1+v^2-2v^2}{2v}\)
\(= \frac{1-v^2}{2v}\)
\(\frac{2v}{1-v^2}dv = \frac{dx}{x}\)
\(\frac{2v}{v^2-1}dv = -\frac{dx}{x}\)
\(\int\frac{2v}{v^2-1}dv = -\int\frac{dx}{x}\)
\(log(v^2-1) = -logx+logc\)
\(log(v^2-1)+ logx = logc\)
\(log(v^2-1)x =logc\)
Substituting \(v=y/x\)
\(log(\frac{y^2}{x^2}-1)x =logc\)
\(log(\frac {y^2-x^2}{x^2}x)=logc\)
\(log(\frac {y^2-x^2}{x})=logc\)
\(y^2-x^2=xc\)
Solution for the differential equation is:
\(y^2-x^2=xc\)
3.2.1.2 QUESTION 2.
Solve the equation \(\frac{dy}{dx}= \frac{y(x-y)}{x^2}\)
Solution
\(\frac{dy}{dx}= \frac{yx}{x^2}-\frac{y^2}{x^2}\)
\(= \frac{yx}{x^2}-(\frac{y}{x})^2\)
\(y =vx\)
\(v+x\frac{dv}{dx}=v-v^2\)
\(x\frac{dv}{dx}=-v^2\)
\(\implies -\frac{1}{v^2}dv=\frac{1}{x}dx\)
\(-\int\frac{1}{v^2}dv=\int\frac{1}{x}dx\)
\(\implies \frac{1}{v} =log(x)+log(c)=log(cx)\)
\(v= \frac{1}{log(cx)}\)
\(v=\frac{y}{x}\)
\(\frac{y}{x}= \frac{1}{log(cx)}\)
\(y= \frac{x}{log(cx)}\)
3.3 Euler’s theorem
Consider a homogeneous function \(f\) of \(x\), \(y\), \(z\),… of degree \(n\), then
\(x\frac{df}{dx} + y\frac{df}{dy} + z\frac{df}{dz} + .... = nf\)
3.3.1 QUESTION 1.
Solve \(f = x^3 + y^3 + 3xy^2\)
Solution
By Euler’s theorem,
\(x\frac{df}{dx} + y\frac{df}{dy} = nf\)
\(f(x,y)\) is a homogeneous function of degree 3
\(\frac{df}{dx}= 3x^2 +3y^2\)
\(\frac{df}{dy}= 3y^2 +6xy\)
Now
\(x(3x^2+3y^2) +y(3y^2+6xy)\)
\(=3x^3+3xy^2+3y^3+6xy^2\)
\(=3x^3+3y^3+9xy^2\)
\(=3(x^3+y^3+3xy^2)\)
\(=3f(x,y)\)