Finding the Integrating Factor

In summary: HallsofIvy's advice is spot on. Equate the two derivatives after differentiating the product of u and the expression preceding the dx and dy. Use the chain rule when differentiating u(xy) with respect to x and y only by denoting product xy as v. Simplify the resulting equation and you'll notice it reduces to a simple separable differential equation. Solve that and express u in terms of x,y only and you're done. I tried it out and it works like a charm.
  • #1
Happydog
3
0
Hi ok so te question is asking for me to fin the integrating factor of (y2x+y)dy + (x2y+2x)dy = 0 the only thing i know is that the integrating factor should be a function of xy

Can some one pleas explain how to do this? Thank you.
 
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  • #2
You appear to have mistyped your problem. There 2 "dy"s here but no "dx".
 
  • #3
Happydog said:
Hi ok so te question is asking for me to fin the integrating factor of (y2x+y)dy + (x2y+2x)dy = 0 the only thing i know is that the integrating factor should be a function of xy

Can some one pleas explain how to do this? Thank you.
Assuming that you mean [itex](y^2x+ y)dx+ (x^2y+ 2x)dy= 0[/itex], if you know that the integrating factor is a function of xy, the obvious thing to do is to put multiply that equation by u(xy) and see what happens. In order that [itex]u(xy)(y^2x+ y)dx+ u(xy)(x^2y+ 2x)dy= 0[/itex] be exact, we must have [itex](u(y^2x+ y))_y[/itex] and [itex](u(x^2y+ 2x)_x)[/itex] equal. That is, [itex]u_x(y^2x+ y)+ u(2xy+1)= u_x(x^2y+ 2x)+ u(2xy+ 2)[/itex]. Since u is a function of xy specifically, [itex]u_x= yu'[/itex] and [itex]u_y= xu'[/itex]. That is, u must satisfy [itex]xu(y^2x+ y)+ u(2xy+1)= yu'(x^2y+ 2x)+ u(2xy+ 2)[/itex]. That gives you a differential equation for u which may or may not be solvable, depending upon whether there really is an integrating factor that is a function of xy only.
 
  • #4
Yes the original equation is (y2x+y)dx + (x2y+2x)dy = 0, i have taken your sugestion but it did not work, i tryed a more general form of xnym and multiplied it through to see if i could solve for the n and m and equate the two of them but the two are not equal i end up getting (m+2)=(n+2) and (m+1)=2(n+1).

so can anyone also give me some advice on how to solve this problem i am having. Thank you.
 
  • #5
I think HallsofIvy's advice is spot on. Equate the two derivatives after differentiating the product of u and the expression preceding the dx and dy. Use the chain rule when differentiating u(xy) with respect to x and y only by denoting product xy as v. Simplify the resulting equation and you'll notice it reduces to a simple separable differential equation. Solve that and express u in terms of x,y only and you're done. I tried it out and it works like a charm.
 
  • #6
Happydog said:
Yes the original equation is (y2x+y)dx + (x2y+2x)dy = 0, i have taken your sugestion but it did not work, i tryed a more general form of xnym and multiplied it through to see if i could solve for the n and m and equate the two of them but the two are not equal i end up getting (m+2)=(n+2) and (m+1)=2(n+1).

so can anyone also give me some advice on how to solve this problem i am having. Thank you.
You told us, originally, that the integrating factor was a function of xy. Why are you now trying xmyn? How do you know the integrating factor is either a function of xy or of the form xmyn?
 
  • #7
Yeah it did work sorry about that, and as for havening it as xmyn other times you can use it to find out what power the x and y are to making it easer. mind you you have to know xy are the integrating factor.
 

What is an integrating factor?

An integrating factor is a function used in solving differential equations. It is multiplied to the equation to transform it into an exact differential equation, making it easier to solve.

Why do we need to find the integrating factor?

We need to find the integrating factor in order to solve certain types of differential equations that cannot be solved using traditional methods. It helps to simplify the equation and make it easier to solve.

How do we find the integrating factor?

The integrating factor can be found by dividing the coefficient of the differential term by the coefficient of the variable term. This will give us the integrating factor, which can then be multiplied to the equation to make it exact.

What are the benefits of using an integrating factor?

Using an integrating factor can make solving certain types of differential equations much easier and more efficient. It can also help to reveal patterns and relationships within the equation that may not have been apparent before.

Are there any limitations to using an integrating factor?

While an integrating factor can be very useful in solving differential equations, it may not work for all types of equations. It is important to understand the limitations and when other methods may be more appropriate.

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