- #1
sourlemon
- 53
- 1
1. Solve the equation
[sin(xy) + xycos(xy)]dx + [1 +x[tex]^{2}[/tex]cos(xy)dy = 0
[sin(xy) + xycos(xy)]dx = M(x,y)
[1 +x[tex]^{2}[/tex]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[tex]^{2}[/tex]ysin(xy)
dN/dy
u = x[tex]^{2}[/tex] du = 2x v= cos(xy) dv = -ysin(xy)
x[tex]^{2}[/tex](-xsin(xy))+ (2x)cos(xy) = (2x)cos(xy) - x[tex]^{2}[/tex]y(sin(xy))
Since dM/dy = dN/dx, the equation is exact.
F(x,y) = [tex]\int[/tex][1 + x[tex]^{2}[/tex]cos(xy)dy + h(x)
dy = y
x[tex]^{2}[/tex]cos(xy)dy = x[tex]^{2}x[/tex]sin(xy) = x[tex]^{3}[/tex]sin(xy)
F(x,y) = y + x[tex]^{3}[/tex]sin(xy) + h(x)
dF/dx(x,y) = N(x,y) = 3x[tex]^{2}[/tex]cos(xy) + h'(x)
N(x,y) = M (x,y)
3x[tex]^{2}[/tex]cos(xy) + h'(x) = sin(xy) + xycos(xy)
h'(x) = sin(xy) + xycos(xy) - 3x[tex]^{2}[/tex]cos(xy)
h(x) = [tex]\int[/tex][sin(xy) + xycos(xy) - 3x[tex]^{2}[/tex]cos(xy)]
sin(xy) = -ycos(xy)
xycos(xy) = uv - [tex]\int[/tex]vdv = xy[tex]^{2}[/tex]sin(xy) - [tex]\int[/tex]ysin(u)(du) = xy[tex]^{2}[/tex]sin(xy) + ycos(xy)
u = xy du = y dv = cos(xy) v=ysin(xy)
3x[tex]^{2}[/tex]cos(xy) = 3x[tex]^{2}[/tex]ysin(xy) - [tex]\int[/tex]ysin(xy)(6x)
u = 3x[tex]^{2}[/tex] du = 6x dv=cos(xy) v = ysin(xy)
I got until there. How do I integrate [tex]\int[/tex]ysin(xy)(6x)? Or did I missed something?
[sin(xy) + xycos(xy)]dx + [1 +x[tex]^{2}[/tex]cos(xy)dy = 0
Homework Equations
The Attempt at a Solution
[sin(xy) + xycos(xy)]dx = M(x,y)
[1 +x[tex]^{2}[/tex]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[tex]^{2}[/tex]ysin(xy)
dN/dy
u = x[tex]^{2}[/tex] du = 2x v= cos(xy) dv = -ysin(xy)
x[tex]^{2}[/tex](-xsin(xy))+ (2x)cos(xy) = (2x)cos(xy) - x[tex]^{2}[/tex]y(sin(xy))
Since dM/dy = dN/dx, the equation is exact.
F(x,y) = [tex]\int[/tex][1 + x[tex]^{2}[/tex]cos(xy)dy + h(x)
dy = y
x[tex]^{2}[/tex]cos(xy)dy = x[tex]^{2}x[/tex]sin(xy) = x[tex]^{3}[/tex]sin(xy)
F(x,y) = y + x[tex]^{3}[/tex]sin(xy) + h(x)
dF/dx(x,y) = N(x,y) = 3x[tex]^{2}[/tex]cos(xy) + h'(x)
N(x,y) = M (x,y)
3x[tex]^{2}[/tex]cos(xy) + h'(x) = sin(xy) + xycos(xy)
h'(x) = sin(xy) + xycos(xy) - 3x[tex]^{2}[/tex]cos(xy)
h(x) = [tex]\int[/tex][sin(xy) + xycos(xy) - 3x[tex]^{2}[/tex]cos(xy)]
sin(xy) = -ycos(xy)
xycos(xy) = uv - [tex]\int[/tex]vdv = xy[tex]^{2}[/tex]sin(xy) - [tex]\int[/tex]ysin(u)(du) = xy[tex]^{2}[/tex]sin(xy) + ycos(xy)
u = xy du = y dv = cos(xy) v=ysin(xy)
3x[tex]^{2}[/tex]cos(xy) = 3x[tex]^{2}[/tex]ysin(xy) - [tex]\int[/tex]ysin(xy)(6x)
u = 3x[tex]^{2}[/tex] du = 6x dv=cos(xy) v = ysin(xy)
I got until there. How do I integrate [tex]\int[/tex]ysin(xy)(6x)? Or did I missed something?