Question

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

Answer 1

`y = 2tan(2x + A) - 2x`

Explanation:

We have:

`dy/dx=(y+2x)^2+2` ..... [A]

We can perform a substitution, Let

`v = y + 2x`

Then if we differentiate, we get:

`(dv)/dx = dy/dx + 2 => dy/dx = (dv)/dx - 2`

Substituting into the original DE `A`, we get:

`(dv)/dx - 2 = v^2+2`
`:. (dv)/dx = v^2+4`

This is now a First Order separable DE, so we can rearrange and seperate the variables, giving:

`int \ 1/(v^2+4) \ dv = int \ dx`

Which we can integrate, using standard integrals, to get:

`1/2arctan (v/2) = x + C`

And, restoring the substitution:

`1/2arctan ((y + 2x)/2) = x + C`
`arctan ((y + 2x)/2) = 2x + 2C`
`:. (y + 2x)/2 = tan(2x + A)`
`:. y+2x = 2tan(2x + A)`
`:. y = 2tan(2x + A) - 2x`