Question

(3xy2 – y)dx – (2x²y– xy²)dy = 0​

Answer 1

Step-by-step explanation:

(3xy² - y³) dx - (2x²y - xy²) dy = 0

⇒ dy/dx = (3xy² - y³)/(2x²y - xy²) ...(i)

Let us take,

y = vx

Then,

dy/dx = v + x (dv/dx)

From (i), we get

v + x (dv/dx) = (3x v²x² - v³x³)/(2x² vx - x v²x²)

⇒ v + x (dv/dx) = (3v² - v³)/(2v - v²)

⇒ x (dv/dx) = (3v - v²)/(2 - v) - v

⇒ x (dv/dx) = (3v - v² - 2v + v²)/(2 - v)

⇒ x (dv/dx) = v/(2 - v)

⇒ {(2 - v)/v} dv = (dx)/x

⇒ 2 (dv)/v - dv = (dx)/x

∴ integrating we get

2 ∫ (dv)/v - ∫ dv = ∫ (dx)/x

⇒ 2 logv - v = logx + c, where c is intergral constant

⇒ 2 log (y/x) - (y/x) = logx + c [ ∵ y = vx ]

⇒ 2 logy - 2 logx - (y/x) = logx + c

⇒ 2 logy - 3 logx = (y/x) + c,

which is the required primitive