Question

Reduce the following differential equation to the variable separable form and hence solve:

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

Answer 1

`("x - y")^2 "dy"/"dx" = "a"^2`     .....(1)

Put x - y = u

∴ x - u = y

∴ 1 - `"du"/"dx" = "dy"/"dx"`

∴ (1) becomes, `"u"^2 (1 - "du"/"dx") = "a"^2`

∴ `"u"^2 - "u"^2 "du"/"dx" = "a"^2`

∴ `"u"^2 - "a"^2 = "u"^2 "du"/"dx"`

∴ dx = `"u"^2/("u"^2 - "a"^2)`du

Integrating both sides, we get

`int "dx" = int (("u"^2 - "a"^2) + "a"^2)/("u"^2 - "a"^2)`du

∴ x = `int 1  "du" + "a"^2 int "du"/("u"^2 - "a"^2) + "c"_1`

`1/"2a" log |("u - a")/("u + a")| + "c"_1`

∴ x = x - y + `"a"/2 log |("x - y - a")/("x - y + a")| + "c"_1`

∴ - c₁ + y = `"a"/2 log |("x - y - a")/("x - y + a")|`

∴ - 2c₁ + 2y =  a log `|("x - y - a")/("x - y + a")|`

∴ c + 2y = a log `|("x - y - a")/("x - y + a")|`, where c = - 2c₁

This is the general solution.