Thread: Exact Differential Equation

I am told to verify that the given differential equation is exact; then solve it:
(x + tan-1y)dx + ((x + y)/(1 + y^2))dy=0

So far I have found the exactness and have gotten down to
F= (x^2) + 2xtan^(-1) (y) + g(y)

I am then trying to find g(y) and to do this do I have to set
F= (x^2) + 2xtan^(-1) (y) + g(y) = ((x + y)/(1 + y^2))
If I do, how in the world do I cancel things out to get the anwer!

The answer is suppose to be
(x^2) + 2xtan^(-1) (y) + ln(1+y^2)

Please help

Thanks

Originally Posted by mandy123

I am told to verify that the given differential equation is exact; then solve it:
(x + tan-1y)dx + ((x + y)/(1 + y^2))dy=0

So far I have found the exactness and have gotten down to
F= (x^2) + 2xtan^(-1) (y) + g(y)

I am then trying to find g(y) and to do this do I have to set
F= (x^2) + 2xtan^(-1) (y) + g(y) = ((x + y)/(1 + y^2))
If I do, how in the world do I cancel things out to get the anwer!

The answer is suppose to be
(x^2) + 2xtan^(-1) (y) + ln(1+y^2)

Please help

Thanks

If is exact then (which you did). If so, then F exist such that



so in your case



From the first we obtain



Now substitute this into the above expression for Fy, i.e.



This will get you an equation for g(y) or g' in particular.