Question

Solve the differential equation `y'=(y-x)^2`?

Answer 1

`y = x+ (a^2+x^2)/(a^2-x^2)`

where `a` is a constant of integration.

Explanation:

Let us change variables to `u = y-x`.

Then `u^' = y^'-1` and so the differential equation becomes

`u^' = u^2-1 implies`

`(du)/(u^2-1) = dx implies`
`1/2 log|(u-1)/(u+1)| = log x -log a`

where we have written the constant of integration as `-log a`. Thus

`(u-1)/(u+1) = x^2/a^2 implies`
`u = (a^2+x^2)/(a^2-x^2) implies`

`y = x+ (a^2+x^2)/(a^2-x^2)`

Answer 2

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

Explanation:

We have:

`y' =(y-x)^2` ..... [A]

Perform the substitution:

`u =y-x`

Then differentiating wrt `x`, and applying the product rule:

`(du)/dx = dy/dx-1 => dy/dx = 1+(du)/dx`

Substituting into the DE:

`1+(du)/dx = u^2`
`:. (du)/dx = u^2-1`

Reducing the DE to a separable form:

`1/(u^2-1) \ (du)/dx = 1`

So we can "separate the variables" to get:

`int \ 1/(u^2-1) \ du = int \ dx`

We can perform partial fraction decomposition of the LHS integrand:

`1/(u^2-1) -= 1/((u+1)(u-1)`
`\ \ \ \ \ \ \ \ \ \ \ \ -= A/(u+1)+B/(u-1)`

And using the "cover up" methods we get:

`A=-1/2, B= 1/2`

Allowing us to write the equation as:

`1/2 int \ 1/(u-1)-1/(u+1) \ du = int \ dx`

And integrating, we get:

`1/2{ln ( u-1) - ln(u+1)} = x + C`

`:. ln ((u-1)/(u+1) ) = 2x + 2C`

`:. (u-1)/(u+1) = e^(2x + 2C)`

`:. u-1 = (u+1)Ae^(2x)`

`:. u-1 = uAe^(2x) + Ae^(2x)`

`:. u - uAe^(2x) = 1 + Ae^(2x)`

`:. u(1 - Ae^(2x)) = 1 + Ae^(2x)`

`:. u = (1 + Ae^(2x))/(1 - Ae^(2x))`

And restoring the substitution:

`y-x = (1 + Ae^(2x))/(1 - Ae^(2x)) => y = x+(1 + Ae^(2x))/(1 - Ae^(2x))`