Finding an Exact Solution for (x+2)sin(y)dx + xcos(y)dy = 0

[Jamin2112]

Homework Statement

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

Homework Equations

I'll bring them in as I solve.

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) = e^xe^ln|x| (Don't need constant e^k)

----------> u(x) = xe^x

----------> ∂/∂y [xe^x(x+2)sin(y)] = ∂/∂x [xe^xxcos(y)]

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

-----------> (x²e^x + 2xe^x)sin(y) + x²e^xcos(y) dy/dx = 0.

Integrate ∂ƒ/∂y with respect to y.

-------------> ƒ(x,y) = -x²e^xsin(y) + g(x)

-------------> ∂f/∂x = -(2x + x²)e^xsin(y) + g'(x) = (x²e^x + 2xe^x)sin(y).

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

Help, please!

 

[vela]

You have the wrong sign on the first term. The integral of cosine is +sine.