Question

What is the general solution of the differential equation `(y^2+1)dy/dx+2xy^2=2x`?

Answer 1

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

Explanation:

We have:

`(y^2+1)dy/dx+2xy^2=2x` ..... [A]

We can rearrange this non-linear First Order differential equation [A] as follows:

`(y^2+1)dy/dx = 2x - 2xy^2`
`:. (y^2+1)dy/dx = -2x(y^2-1)`
`:. (y^2+1)/(y^2-1)dy/dx = -2x`

This is now separable, so we can "seperate the variables" to get:

`int \ (y^2+1)/(y^2-1) \ dy = int \ -2x \ dx` ..... [B]

The RHS integral is standard, and the LHS will require a little manipulation, as follows:

`int \ (t^2+1)/(t^2-1) \ dt = int \ (t^2-1+2)/(t^2-1) \ dt`
`" " = int \ 1 + 2/(t^2-1) \ dt`
`" " = int \ 1 + 2/((t+1)(t-1)) \ dt`

We can now decompose the fractional part of the integrand into partial fractions, as follows:

`2/((t+1)(t-1)) -= A/(t+1) + B/(t-1)`
`" " = ( A(t-1) + B(t+1) ) / ( (t+1)(t-1) )`

Leading to the identity:

`2 -= A(t-1) + B(t+1)`

Where `A,B` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `t = -1 => 2 = -2A => A = -1`
Put `t = +1 => 2 = +2B => B = +1`

So using partial fraction decomposition we have:

`int \ (t^2+1)/(t^2-1) \ dt = int \ 1 - 1/(t+1) + 1/(t-1) \ dt`

Using this result we can now integrate [B] as follows:

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

`:. int \ -1 + 1/(y+1) - 1/(y-1) \ dy = int \ 2x \ dx`

`:. - y + ln| y+1 | - ln| y-1 | = x^2 + C`

`:. ln| (y+1)/(y-1) | -y = x^2 + C`

Which, is the General Solution .

We are unable to find a particular solution, as requested, as noi initial conditions have been provided to allow the constant `C` to be evaluated.