Question

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

Answer 1

`" The G.S. is "4y=3x^2+c.`

Explanation:

`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.`

Answer 2

`y = Ae^(3/4x^2)`

Explanation:

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