Question

Solve: (x + y)(dx – dy) = dx + dy. [Hint: Substitute x + y = z after seperating dx and dy]

Answer 1

Given differential equation is (x + y)(dx – dy) = dx + dy

⇒ (x + y) dx – (x + y) dy = dx + dy

⇒ – (x + y) dy – dy = dx – (x + y)dx

⇒ – (x + y + 1) dy = – (x + y – 1)dx

⇒ `"dy"/"dx" = (x + y - 1)/(x + y + 1)`

Put x + y = z

∴ `1 + "dy"/"dx" = "dz"/"dx"`

`"dy"/"dx" = "dz"/"dx" - 1`

So, `"dz"/"dx" - 1 = (z - 1)/(z + 1)`

⇒ `"dz"/"dx" = (z - 1)/(z + 1) + 1`

⇒ `"dz"/"dx" = (z - 1 + z + 1)/(z + 1)`

⇒ `"dz"/"dx" = (2z)/(z + 1)`

⇒ `(z + 1)/z "d"z` = 2 . dx

Integrating both sides, we get

`int (z + 1)/z "d"z = 2 int "d"x`

⇒ `int(1 + 1/2) "d"z = 2int "d"x`

⇒ `z + log|z| = 2x + log|"c"|`

⇒ `x + y + log|x + y| = 2x + log|"c"|`

⇒ `y + log|x + y| = x + log |"c"|`

⇒ `log|x + y| = x - y + log|"c"|`

⇒ `log|x + y| - log|"c"| = (x - y)`

⇒ `log|(x + y)/"c"| = (x - y)`

⇒ `(x + y)/"c" = "e"^(x - y)`

∴ x + y = `"c" . "e"^(x - y)`

Hence, the required solution is x + y = `"c" . "e"^(x - y)`.