What is the general solution of the differential equation? # (3x^2+2y)dx+2xdy=0 #

2 Answers
Dec 30, 2016

# y = -1/2x^2 + A/x #

Explanation:

We can write the equation # (3x^2+2y)dx+2xdy=0 # as:

# \ \ \ \ \ \ \ 2xdy = -(3x^2+2y)dx #
# :. 2xdy/dx = -3x^2 - 2y #
# :. dy/dx + y/x = -3/2x #

This is a First Order DE of the form:

# y'(x) + P(x)y = Q(x) #

Which we know how to solve using an Integrating Factor given by:

# IF = e^(int P(x) \ dx) #

And so our Integrating Factor is:

# IF = e^(int 1/x \ dx) #
# \ \ \ \ = e^(ln|x|) #
# \ \ \ \ = x #

If we multiply by this Integrating Factor we will (by its very design) have the perfect differential of a product:

# \ \ \ \ \ dy/dx + y/x = -3/2x #

# :. xdy/dx +y = -3/2x^2 #

# :. \ \ d/dx(xy) = -3/2x^2 #

Which is now a separable DE, and we can separate the variables to get:

# " "xy = int \ -3/2x^2 \ dx #
# :. xy = -1/2x^3 + A #
# :. y = -1/2x^2 + A/x #

Dec 30, 2016

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

Explanation:

An alternative approach

The equation is exact. ie there exists a potential function #f(x,y) = C# such that #df = f_x dx + f_y dy = 0#

to verify this we check the mixed partials

#f_x = 3x^2 + 2y implies f_(xy) = 2#

#f_y = 2x implies f_(yx) = 2 qquad star#

# f_(xy) = f_(yx) # !!!

For reasons that become apparent, we can now choose to integrate either term ... we will integrate #f_x# wrt x

#f_x = 3x^2 + 2y implies f = x^3 + 2xy + alpha(y)#

....where #alpha(y)# is the function in y only that may have been lopped off by a differentiation wrt x.

We then look at the other partial that follows by now differentiating this f wrt y:

#f = x^3 + 2xy + alpha(y) implies f_y = 2x + alpha'(y)#

And from #star#

#alpha'(y) = 0 implies alpha(y) = C#

#implies f = x^3 + 2xy + C = tilde C#

Final bit of algebra:

#x(x^2 + 2y) = C#

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

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