Question

Solve the following differential equation:

y dx + (x - y²) dy = 0

Answer 1

y dx + (x - y²) dy = 0

∴ y dx = - (x - y²) dy

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

∴ `"dx"/"dy" + (1/"y") * "x" = "y"`     ....(1)

This is the linear differential equation of the form

`"dx"/"dy" + "P" * "x" = "Q"`, where P = `1/"y"` and Q = y

∴ I.F. = `"e"^(int "P dy") = "e"^(int 1/"y" "dy") = "e"^(log "y") = "y"`

∴ the solution of (1) is given by

`"x" * ("I.F.") = int "Q" * (I.F.) "dy" + "c"_1`

∴ `"xy" = int "y" * "y"  "dy" + "c"_1`

∴ `"xy" = int "y"^2 "dy" + "c"_1`

∴ `"xy" = "y"^3/3 + "c"_1`

∴ `"y"^3/3 = "xy" + "c"`, where c = -c₁ 

This is the general solution.