Question

Integrating factor for differentiation equations y(1+xy) dx +(1+xy+x²y²)dy=0 ​

Answer 1

Answer:

-1

Step-by-step explanation:

The given differential equation is

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

(ydx + x dy) + xy² dx - x²ydy = 0

ydx+xdy+¹dx-dy = 0

X

fydx + xy +f¹dx - f¹dy = 0

...(1)

Let | = fydx + x dy

Put x y = t so that x dy + y dx = dt = √²/1/dt = √₁-² dt = 1/² 1 ху

.. from (1),

1 xy + logx-logy = c

Now y(1) = 1

y=1 when x = 1

+log 1 log 1 = c → C = -1 from (1), solution of differential equation is + log x-logy = -1