What is the general solution of the differential equation #2dy = 3xy \ dx#?

#2dy=3xydx rArr 2dy/y=3xdx,# which is a Separable Variable

Type Diff. Eqn.

To find its General Soln. (G.S.), we integrate it term-wise,

#int2dy=int3xdx+C.#

#:. 2y=3(x^2/2)+C, or, 4y=3x^2+c, c=2C.#

We have:

#2dy = 3xy \ dx#

We should really write this as (because #d/dx# is an operator not a fraction):

#2dy/dx = 3xy #

Method 1 - Separating the Variables

If collect terms in #y# and #x# respectively:

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

And then we "separate the variables" to get:

# int \ 2/y \ dy = int\ 3x \ dx#

Integrating we get:

# \ \ \ 2 ln y = (3x^2)/2 + K #
# :. ln y = 3x^4 + 1/2K #
# :. \ \ \ \ y = e^(3/4x^2 + 1/2K) #
# :. \ \ \ \ y = e^(3/4x^2)e^(1/2K) #
# :. \ \ \ \ y = Ae^(3/4x^2) #

Method 2 - Integrating Factor

W can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

So:

# 2dy/dx = 3xy #

# :. dy/dx - 3/2xy = 0 # ..... [1]

Then the integrating factor is given by;

# I = e^(int P(x) dx) #
# \ \ = exp(int \ -3/2x \ dx) #
# \ \ = exp( -3/4x^2 ) #
# \ \ = e^(-3/4x^2) #

And if we multiply the DE [1] by this Integrating Factor, #I#, we will have a perfect product differential;

# dy/dx - 3/2xy = 0 #

# :. e^(-3/4x^2)dy/dx - 3/2xe^(-3/4x^2)y = 0 #

# :. d/dx {e^(-3/4x^2)y} = 0 #

Which we can directly integrate to get:

# e^(-3/4x^2)y = A #

And multiplying by #e^(3/4x^2)# gives:

# y = A ^(3/4x^2) #, as above