# Homework Help: Special Integrating Factor?

1. Sep 20, 2007

### sourlemon

1. Solve the equation
[sin(xy) + xycos(xy)]dx + [1 +x$$^{2}$$cos(xy)dy = 0

2. Relevant equations

3. The attempt at a solution
[sin(xy) + xycos(xy)]dx = M(x,y)
[1 +x$$^{2}$$cos(xy)dy = N(x,y)

dM/dy
sin(xy) = xcos(xy)

xycos(xy) = uv' + vu' = (xy)(-xsin(xy)) + (x)(cos(xy))
u = xy du = x v = cos(xy) dv=-xsin(xy)

dM/dy= xcos(xy) + (xy)(-xsin(xy)) + (x)(cos(xy)) = 2xcos(xy) - x$$^{2}$$ysin(xy)

dN/dy
u = x$$^{2}$$ du = 2x v= cos(xy) dv = -ysin(xy)
x$$^{2}$$(-xsin(xy))+ (2x)cos(xy) = (2x)cos(xy) - x$$^{2}$$y(sin(xy))

Since dM/dy = dN/dx, the equation is exact.

F(x,y) = $$\int$$[1 + x$$^{2}$$cos(xy)dy + h(x)
dy = y
x$$^{2}$$cos(xy)dy = x$$^{2}x$$sin(xy) = x$$^{3}$$sin(xy)

F(x,y) = y + x$$^{3}$$sin(xy) + h(x)

dF/dx(x,y) = N(x,y) = 3x$$^{2}$$cos(xy) + h'(x)

N(x,y) = M (x,y)

3x$$^{2}$$cos(xy) + h'(x) = sin(xy) + xycos(xy)

h'(x) = sin(xy) + xycos(xy) - 3x$$^{2}$$cos(xy)

h(x) = $$\int$$[sin(xy) + xycos(xy) - 3x$$^{2}$$cos(xy)]

sin(xy) = -ycos(xy)

xycos(xy) = uv - $$\int$$vdv = xy$$^{2}$$sin(xy) - $$\int$$ysin(u)(du) = xy$$^{2}$$sin(xy) + ycos(xy)
u = xy du = y dv = cos(xy) v=ysin(xy)

3x$$^{2}$$cos(xy) = 3x$$^{2}$$ysin(xy) - $$\int$$ysin(xy)(6x)
u = 3x$$^{2}$$ du = 6x dv=cos(xy) v = ysin(xy)

I got until there. How do I integrate $$\int$$ysin(xy)(6x)? Or did I missed something?

2. Sep 20, 2007

### Dick

You are doing some pretty odd differentiating. For the common derivative of M and N, I get 2xcos(xy)-x^3*sin(xy). And when I integrate x^2*cos(xy)dy I get x*sin(xy). I sort of looks like you are mixing up pieces of differentiation and integration.

3. Sep 20, 2007

### sourlemon

:(

How did you get (x^3)(sinxy)?

xycos(xy)

u = xy du = x v = cos(xy) dv=-xsin(xy)

= uv' + vu' = (xy)(-xsin(xy)) + (x)(cos(xy))

I still have this 2xcos(xy) - x^2ysin(xy) :(

4. Sep 20, 2007

### Dick

OOOOPS. Apologies, I sent the wrong thing. You are right on the differentiation. But I still don't like the integral of x^2*cos(xy)dy. Can you check that one again. I still get x*sin(xy).

5. Sep 20, 2007

### sourlemon

Unfortunately, the integral is correct.

I guess I should switch to M(x,y).

May I ask how to ingrate $$\int$$[sin(xy)dx?

I want to set xy = u, which means du = y, but I don't have du.

6. Sep 20, 2007

### Dick

d/dy(x*sin(xy))=x^2*cos(xy). d/dy(x^3*sin(xy))=x^4*cos(xy). x^3*sin(xy) is not the correct antiderivative! To do the integral, sure u=xy, du=y*dx, dx=du/y. So integrate sin(u)*du/y (y is constant this time). So I get -cos(u)/y=-cos(xy)/y.