Question

What is a solution to the differential equation `dy/dx=(2y+x^2)/x`?

Answer 1

`y(x) = (C+log_e(x))x^2`

Explanation:

This is a linear differential equation. Its solution can be composed of the homogeneous solution (`y_h(x)`) plus a particular solution (`y_p(x)`) so

`y(x) = y_h(x) + y_p(x)`

The homogeneous solution obeys

`(dy_h)/(dx) -2/x y_h(x) = 0`. Here, clearly the solution is

`y_h(x) = C x^2`

By grouping variables:
`dy_h/(y_h)=2dx/x->ln y_h=C_1lnx^2->y_h=Cx^2`

The determination of `y_p(x)` must obey

`(dy_p)/(dx)-2/x y_p(x) = x`.

To solve this, we will use a technique due to Lagrange
(https://en.wikipedia.org/wiki/Joseph-Louis_de_Lagrange)

called the constant variation technique.

We will suppose that `y_p(x) = c(x)y_h(x)`. Substituting we obtain

`y_h(x)(dc(x))/(dx) + c(x) ((dy_h(x))/(dx)-(2 y_h(x))/x) = x` or

`y_h(x)(dc(x))/(dx) = x` so

`(dc(x))/(dx) = x/(y_h(x)) = x/( x^2) = 1/x` so

`c(x)= log_e x` and finally

`y(x) = C x^2+log_e(x) x^2 = (C+log_e(x))x^2`

Answer 2

`y= x^2 (ln x + C)`

Explanation:

Alternative Approach

`y'=(2y+x^2)/x = 2 y/x + x`

`y'- 2/x y = x`

This is exact if we times it by a factor of `1/x^2` [See below for mechanical process to find that factor]

`1/x^2 y'- 2/x^3 y = 1/x`

ie
`(1/x^2 y)'= 1/x`

`1/x^2 y= ln x + C`

`y= x^2 (ln x + C)`

Integrating Factor
We can extract the "integrating factor" `eta(x)` for the DE in form `y' color(red)(- 2/x) y = x` as:

`eta = exp (int color(red)( -2/x) dx )`

`= e^(-2 ln x )`

`= 1/x^2`