Help proving a differential equation solution

[ballstix]

It's the first problem of my assignment, and I have no idea what I'm doing:

Show that y = x - x^-1 is a solution for the differential equation xy' + y = 2x.

I tried saying that y' = (2x - y) / x, but I honestly don't even know what to do from there, or if that's even a correct approach.

 

[Deleted member 4993]

It's the first problem of my assignment, and I have no idea what I'm doing:

Show that y = x - x^-1 is a solution for the differential equation xy' + y = 2x.

I tried saying that

y' = (2x - y) / x, <<< How did you get that?

Derive y' from the given y [y =x - x^(-1)]

but I honestly don't even know what to do from there, or if that's even a correct approach.

Now simply replace y' (to the left-hand-side) in the given ODE - and show that it reduces to '2x'.

 

[ballstix]

Oh, sorry, I got that equation (y' = (2x - y) / x) by solving for y' using the given differential equation.

Thanks; I think I'm just confused about how you're supposed to deal with having both x and y in the equation for y'.

 

[Deleted member 4993]

Oh, sorry, I got that equation (y' = (2x - y) / x) by solving for y' using the given differential equation.

Thanks; I think I'm just confused about how you're supposed to deal with having both x and y in the equation for y'.

If you derive y' as I have suggested (and that is the correct procedure) - you won't have that probelm (of having 'x' and 'y' in the expression for y').

 

[ballstix]

Okay, so I take the derivative and end up with y' = 1 - (1 / -x^2). Then I plug that into the other equation: x(1 - (x / -x^2)) + y = 2x, right? How do I get it down to 2x with that y in there? I'm sorry if I seem so clueless.

Wait, just kidding. Thanks for your help.