Question

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

(2x - 2y + 3)dx - (x - y + 1)dy = 0, when x = 0, y = 1.

Answer 1

(2x - 2y + 3)dx - (x - y + 1)dy = 0

∴ (x - y + 1)dy = (2x - 2y + 3) dx

∴ `"dy"/"dx" = (2("x - y" + 3))/(("x - y") + 1)`   ....(1)

Put x - y = u. Then `1 - "dy"/"dx" = "du"/"dx"`

∴ `"dy"/"dx" = 1 - "du"/"dx"`

∴ (1) becomes, `1 - "du"/"dx" = (2"u" + 3)/("u" + 1)` 

∴ `"du"/"dx" = 1 - (2"u" + 3)/("u" + 1) = ("u" + 1 - 2"u" - 3)/("u + 1")`

∴ `"du"/"dx" = (- "u" - 2)/("u" + 1) = - (("u + 2")/("u + 1"))`

∴ `("u + 1")/("u + 2")`du = - dx

Integrating both sides, we get

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

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

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

∴ u - log |u + 2| = - x + c

∴ x - y - log |x - y + 2| = - x + c

∴ (2x - y) - log |x - y + 2| = c

This is the general solution.

Now, y = 1, when x = 0

∴ (0 - 1) - log |0 - 1 + 2| = c

∴ - 1 - 0 = c

∴ c = - 1

∴ the particular solution is

(2x - y) - log |x - y + 2| = - 1

∴ (2x - y) - log |x - y + 2| + 1 = 0