Solve using variable separable

Question

(y + 1) dx - (xy + x - 3y - 3) dy = 0, when x = 5, y = ln 4

a. y = (x - 3)^2

b. y = ln (2x - 6)

c. y = ln (x - 3)^2

d. y = (2 - cos t)^-3

Answer 1

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

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

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

dy = (1/(x-3))dx

∫dy = ∫(1/(x-3))dx

y = ln lx-3l + C

When x = 5 and y = ln4, ln4 = ln2 + C So, C = ln4 - ln2 = ln(4/2) = ln2.

y = lnlx-3l + ln2

y = ln(2lx-3l)= ln(2x - 6), when x > 3.