Special Integrating Factor?

Yeah, it does not look like your step of having u=xy was productive. Just integrate sin(xy)dx? There is no y in there. In summary, the conversation covers solving an equation by finding the derivative of M and N, determining that the equation is exact, and finding the antiderivative by setting u=xy and using integration by parts. There is also a minor discrepancy in the integral of x^2*cos(xy)dy.
  • #1
sourlemon
53
1
1. Solve the equation
[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?
 
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  • #2
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
:(

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
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
Unfortunately, the integral is correct.

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

May I ask how to ingrate [tex]\int[/tex][sin(xy)dx?

I want to set xy = u, which means du = y, but I don't have du.
 
  • #6
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.
 

1. What is a special integrating factor?

A special integrating factor is a function that is multiplied to a differential equation to make it easier to solve. It is used in cases where the integrating factor is not constant or not easily identifiable.

2. How is a special integrating factor determined?

A special integrating factor is determined by finding a function that, when multiplied to the original differential equation, makes the left side of the equation a total derivative of the right side. This can be done through various methods such as inspection, substitution, or trial and error.

3. Why is a special integrating factor useful?

A special integrating factor is useful because it simplifies the process of solving a differential equation. It can transform a complex or non-linear equation into a more manageable form, making it easier to find a solution.

4. Are there any limitations to using a special integrating factor?

Yes, there are limitations to using a special integrating factor. It may not always be possible to find a suitable integrating factor for a given differential equation. In addition, the use of a special integrating factor may result in a more complex solution, which can be more difficult to interpret.

5. Can a special integrating factor be used for any type of differential equation?

No, a special integrating factor is typically used for first-order linear differential equations. It may also be used for certain non-linear equations, but it is not applicable to all types of differential equations.

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