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

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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#