Solve the differential equation. 1+xy=xy'. I got an answer but not sure about it.

Question

What is the solution?

Answer 1

I suggest using the integrating factor method.

 

1 + xy = xy'

 

xy' - xy = 1

 

y' - y = 1/x

 

This has the form 

 

y' + P(x)y = Q(x), where

 

P(x) = -1, Q(x) = 1/x

 

Integrating factor:

 

e^∫P(x)dx = e^∫-dx = e^-x

 

Using an integrating factor, the general solution for y' + P(x)y = Q(x) is

 

y = e^-∫P(x)dx [∫e^∫P(x)dx Q(x)dx + c]

 

y = e^x [∫(e^-x/x)dx + c]

 

This gives the solution in terms of an integral.  Note that if you substitute this back into

 

y' - y = 1/x, it satisfies the equation.

 

Does this help?