How the integrating factor of (ydx-xdy)=0 is (-1/(x^2+y^2))

[math_worm]

Can anyone help in finding the integrating factor of (ydx-xdy)=0

According Schaum's outline of differential equations 3rd edition

it's integrating factor are (-1/x^2), (-1/(x^2+y^2)).

plz help me.

May God bless those who will help me.

[K!!]

Why do you need an integrating factor? Your equation is separable.

[Country Boy]

K!! is right, that equation is separable and integrates to give y= Cx. But that doesn't mean you can't look for an integrating factor!

 

There is no general method of finding integrating factors. One thing you can do is assume some particular "type" of function and just try that.

 

For example, if I look for an integrating factor that is a function of x only, say v(x), multiplying by v(x) we have v(x)ydx- v(x)xdy. In order that this be "exact", we must have (v(x)y)_y= (-v(x)x)_x. On the left, we have (v(x)y)_y= v(x), on the right, (-v(x)x)_x= -v'x- v. Setting those equal, v= -xv'- v or xv'= -2v. That is, again, a separable equation: dv/v= -2dx/x which integrates to ln(v)= 2ln(x)= ln(x^(-2)) so v= x^(-2)= 1/x^2.

 

How did I know to look for an integrating factor that was a function of x only? Lucky guess. (Helped by the fact that you had already told me that 1/x^2 was an integrating factor!)

 

From looking at ydx- xdy, you might just as well guess a function of y only: v(y). Then we would have v(y)ydx- v(y)xdy and would have to have yv'+ v= -v so that yv'= -2v. Yes, v(y)= 1/y^2 is also a perfectly good integrating factor.

 

The form of -1/(x^2+ y^2) makes me suspect that they changed to polar coordinates and found the integrating factor from that.

 

If x= r cos(t), y= r sin(t), then dx= cos(t)dr- r sin(t)dt and dy= sin(t)dr+ r cos(t)dt.

 

ydx= (r sin(t))(cos(t)dr)- (r sin(t))(r sin(t)dt)

= r sin(t)cos(t)dr- r^2 sin^2(t)dt

- xdy= -(r cos(t))(sin9t)dr)- (r cos(t))(r cos(t)dt)

= -r sin(t)cos(t)dr- r^2 cos^2(t)dt

 

so ydx- xdy= -r^2 dt. Since the coefficient of dr is 0, the derivative of that multiplied by anything, with respect to t, will be 0. In order to have this exact, we must also have the derivative of the coefficient of dt be 0 which is true, of course, if that coefficient is a constant: multiplying by 1/r^2 give -1 dt. An integrating factor is 1/r^2= 1/(x^2+ y^2). (The "-1" in the numerator of what you give is irrelevant).

 

Notice that there are many possible integrating factors- and no general method for finding them.