# Exact Differential Equation

• Feb 4th 2009, 01:30 PM
mandy123
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)

Thanks
• Feb 4th 2009, 03:00 PM
Jester
Quote:

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)

Thanks

If $\displaystyle M dx + N dy = 0$ is exact then $\displaystyle M_y = N_x$ (which you did). If so, then F exist such that

$\displaystyle F_x = m,\;\;\;F_y = N$

so in your case

$\displaystyle F_x = x + \tan^{-1}y,\;\;\;F_y = \frac{x+y}{1+y^2}$

From the first we obtain

$\displaystyle F = \frac{x^2}{2} + x\tan^{-1}y + g(y)$

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

$\displaystyle F_y = \frac{x+y}{1+y^2}$

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