Question

Which of the following functions is an integrating factor for the differential equation ydx - xdy = 0)

(a)-1/x²
(b)1/y²
(c)1/xy
(d) All of these
​

Answer 1

Given that

  ydx - xdy =0

The general form Mdx + Ndy =0

  And  du =$\frac{du}{dx}$ dx + $\frac{du}{dy}$ dy

$\frac{d}{dy}$ $\frac{x}{y}$  = y dx/dy - x dy/dy / $y^{2}$

d(x/y)  = y dx - xdy /$y^{2}$ =0

$\frac{1}{y^{2} }$  = Integrating factor

d(x/y) =du=0

u= x/y

u = c

c= x/y

y= cx

The answer is all of these

Answer 2

Answer:

1/$y^{2}$

Step-by-step explanation:

Integrating factor of ydx-xdy=0

We know, d/dy(x/y) = (y(dx/dy) - x(dy/dy))/ y²

Multiplying both sides by dy we get,

d(x/y) = (y dx - x dy) /y² = 0

From the above equation we get 1/y² is an integrating factor.

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