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