# Finding an exact solution

1. Apr 15, 2010

### Jamin2112

1. The problem statement, all variables and given/known data

Solve (x+2)sin(y)dx + xcos(y)dy = 0 by finding an integrating factor u(x).

2. Relevant equations

I'll bring them in as I solve.

3. The attempt at a solution

I assume there's some parent function ƒ(x,y)=c.

---> ∂ƒ/∂x, ∂ƒ/∂y=0

Now I'll just put it into the form M(x,y) + N(x,y)dy/dx = 0.

(x+2)sin(y) + xcos(y)dy/dx = 0.

An integrating factor is just some multiplied by the equation that helps us solve it. In this case I want ∂M/∂y=∂N/∂x, so I'll just modify M and N by multiplying them by a function of x (because the problem says to use u(x)).

∂/∂y [u(x)(x+2)sin(y)] = ∂/∂x [u(x)xcos(y)]

-----> u(x)(x+2)cos(y) = [u(x)*1+x*u'(x)]cos(y) (product rule used on x terms)

------> u(x)(x+2)=u(x)+x*u'(x)

------>u(x)(x+1)=x*u'(x)

------> u'(x)/u(x) = 1 + 1/x

------> d/dx ln|u(x)| = 1 + 1/x

-------> ln|u(x)| = x + ln|x| + K

--------> u(x) = exeln|x| (Don't need constant ek)

----------> u(x) = xex

----------> ∂/∂y [xex(x+2)sin(y)] = ∂/∂x [xexxcos(y)]

-----------> ∂M/∂y=∂N/∂x. Mission Accomplished. Almost.

-----------> (x2ex + 2xex)sin(y) + x2excos(y) dy/dx = 0.

Integrate ∂ƒ/∂y with respect to y.

-------------> ƒ(x,y) = -x2exsin(y) + g(x)

-------------> ∂f/∂x = -(2x + x2)exsin(y) + g'(x) = (x2ex + 2xex)sin(y).

But wait! This is supposed to cancel out the y terms; it doesn't.