What is the general solution of the differential equation? # y^2 \ dx + xy \ dy = 0 #

What can you say about the solutions?

2 Answers
Dec 10, 2017

Solution of this differential euation was #1/2x^2y^2=C#

Explanation:

#M(x, y)=y^2# and #N(x,y)=xy#

Once, I controlled condition exactness of differential equation,

#(dM)/dy=2y# and #(dN)/dx=y#

Hence, it wasn't exact.

I decided to integration factor accordingly #x#

Hence,

#P=e^[int ((dM)/dy-(dN)/dx)/N*dx#

=#e^[int (2y-y)/(xy)*dx#

=#e^[int (dx)/x]#

=#e^(Lnx)#

=#x#

After multiplying both sides with #P=x#,

#xy^2*dx+x^2y*dy=0#

After controlling exactness condition of it,

#(d(xy^2))/dy=(d(x^2y))/dx=2xy#

Thus, solution of it,

#xy^2*dx+x^2y*dy=0#

#d(1/2x^2y^2)=0#

#1/2x^2y^2=C#

Dec 11, 2017

Multiplying by #x# we get an exact equation:

# 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-N_x=0#

# 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 #mu(u)# such that

# (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# and #N#:

# 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_y = N#, then:

# 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 = M# and we differentiate the last result to get:

# f_x = xy^2 + g'(x) #

As #f_x=M# then we have:

# 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#