Question

What is the solution of the Differential Equation `dy/dx=(x-3)y^2/x^3`?

Answer 1

`y = (2x^2)/(2x-3+Ax^2)`

Explanation:

We have:

`dy/dx=(x-3)y^2/x^3`

This is a first Order non-linear Separable Differential Equation, we can collect terms by rearranging the equation as follows

`1/y^2 dy/dx=(x-3)/x^3`

And now we can "separate the variables" to get

`int \ 1/y^2 \ dy= int \ (x-3)/x^3 \ dx`

`:. int \ 1/y^2 \ dy= int \ 1/x^2-3/x^3 \ dx`

And integrating gives us:

`y^(-1)/(-1) = x^(-1)/(-1)-3x^(-2)/(-2) + C_1`

`:. -1/y = -1/x+3/(2x^2) + C_1`

`:. -1/y = (3+2C_1x^2-2x)/(2x^2)`

`:. 1/y = (2x-3+Ax^2)/(2x^2)`

`y = (2x^2)/(2x-3+Ax^2)`

Answer 2

`-1/y=-1/x+3/(2x^2)+C`

Explanation:

Use separation of variables, that is put the `y` terms in the left, and `x` terms in the right.

`dy/(dx)=(x-3)y^2/x^3`

`(1/y^2)dy/(dx)=(x-3)/x^3`

`(1/y^2)dy=(x-3)/x^3dx`

Integrate both sides:

`int (1/y^2)dy=int (x-3)/x^3dx`

`-1/y=int1/x^2-3/x^3dx`

`-1/y=-1/x-(-3/(2x^2))+C`

`-1/y=-1/x+3/(2x^2)+C`