The degree of the product differentiation represents the operations in the differentiation which deals with the product of the terms. For example if we have the function like xy then the degree of the function is 2 [x^1y^1 = ^2 ]. The homogeneous equation represents the degreee of the functions in the numerator and the denominator are same. In this article we are going to discuss about the degree of product differentiation for the students in detail with the homogeneous equations in different form with clear view.
Examples for the degree of product differentiation
Review on the degree of product differentiation for the differentiable function dy/dx = [y^2-x^2]/[2xy]
Solution:
The function given for differentiation is dy/dx = [y^2-x^2]/[2xy]
Here it is present in the homogeneous form.
Since the numerator degree[ [y^2-x^2]] is two and the denominator [ [2x^1y^1]] degree is also two
Put y = vx and dy/dx = v + xdv/dx in the equation given we get,
=> v + x[dv]/dx = [(vx)^2-x^2]/[2x(vx)]
=> v + x[dv]/dx = [v^2x^2-x^2]/[2vx^2]
=> v + x[dv]/dx = x^2[v^2-1]/[2vx^2]
=> v + x[dv]/dx = [v^2-1]/[2v]
=> x[dv]/dx = [v^2-1]/[2v] - v
=> x[dv]/dx = [v^2-1-2v^2]/[2v]
=> x[dv]/dx = [-1-v^2]/[2v]
=> [2v]/[-[1+v^2]] dv = dx/x
=> [2v]/[1+v^2] dv = -1/x dx
Apply integration on both sides, we get
=> int [2v]/[1+v^2] dv = int -1/x dx
=> int [2v]/[1+v^2] dv = -int 1/x dx
=> log|1 + v^2| = - log |x| + log c
=> log|1 + v^2| + log |x| = log c
=> log|x[1 + v^2]| = log c
=> x[1 + v^2] = +- c
Put y = vx and v = y/x
=> x[1 + [y/x]^2] = C_1
=> x[1 + y^2/x^2] = C_1
=> x[x^2 + y^2]/x^2 = C_1
=> [x^2 + y^2]/x = C_1
=> x^2 + y^2 = xC_1 is the answer for the differentiation for the degree of product.
Problems for the degree of product differentiation
Review on the degree of product differentiation for the differentiable function dy/dx = [x^2y]/[x^3 + y^3 ]
Solution:
The answer is [-x^3]/[3y^3] + log|y| = C
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