What is the general solution of the differential equation? # y^2 \ dx + xy \ dy = 0 #
What can you say about the solutions?
What can you say about the solutions?
2 Answers
Solution of this differential euation was
Explanation:
Once, I controlled condition exactness of differential equation,
Hence, it wasn't exact.
I decided to integration factor accordingly
Hence,
=
=
=
=
After multiplying both sides with
After controlling exactness condition of it,
Thus, solution of it,
Multiplying by
# xy^2 \ dx + x^2y \ dy = 0 #
Whose solution is:
# y = +- C/x#
ie a family of hyperbolas.
Explanation:
# y^2 \ dx + xy \ dy = 0 ..... [A]#
Suppose we have:
# M(x,y) \ dx + N(x,y) \ dy = 0#
Then the DE is exact if
# M = y^2 => M_y = 2y #
# N= xy => N_x = y #
# M_y - N_x != 0 => # Not an exact DE
So, we seek an Integrating Factor
# (muM)_y = (muN)_x#
So, we compute::
# I=(M_y-N_x)/N = (2y-y)/(xy) = 1/x #
So the Integrating Factor is given by:
# mu(x) = e^(int \ I \ dx) #
# \ \ \ \ \ \ \ = e^(int \ 1/x \ dx) #
# \ \ \ \ \ \ \ = e^(lnx) #
# \ \ \ \ \ \ \ = x #
So when we multiply the DE [A] by the IF we now get an exact equation:
# (x)y^2 \ dx + (x)xy \ dy = 0 .#
# xy^2 \ dx + x^2y \ dy = 0 #
And so if we redefine the function
# M = xy^2 => M_y = 2xy #
# N= x^2y => N_x = 2xy #
Making the equation exact
Then, our solution is given by:
# f_x = M# and#f_y = N# and
If we consider
# f = int \ x^2y \ partial y + g(x) # , where we treat#x# as constant
# \ \ = (x^2y^2)/2 + g(x) #
And now we consider
# f_x = xy^2 + g'(x) #
As
# xy^2 + g'(x) = xy^2 #
# :. g'(x) = 0 => g(x) = K#
Leading to the GS:
# (x^2y^2)/2 = K => (xy)^2=c #
# :. xy=+- C => y = +- C/x#