Exact Differential Equations and Integrating Factor

In summary, the problem is to show that the given differential equation is exact, and then find its general solution using integrating factors. The equation is M(x,y)dx + N(x,y)dy = 0, where M(x,y) = (x2 + y2 + x) and N(x,y) = xy. To show exactness, the integrating factor μ(x) = e∫h(x)dx is used, where h(x) = (2y - y) / xy = 1/x. The integrating factor is then found to be x. The solution for the general solution is 3x4 + 6x2y2 + 4x3 = C. However, the process and method
  • #1
Color_of_Cyan
386
0

Homework Statement


"Show that each of the given differential equations of the form M(x,y)dx + N(x,y)dy = 0 are
exact, and then find their general solution using integrating factors μ(x) = e∫h(x)dx and μ(x) = e∫g(y)dy

Homework Equations


(x2 + y2 + x)dx + (xy)dy = 0

The Attempt at a Solution

Can someone please tell me how to get started or what to do after? This is very confusing for me. I did THIS EXACT SAME problem before but in another way, where it said to show exactness and I had to use the integrating factor to show exactness and THEN solve using grouping. NOW it's saying to use integrating factor to find the general solution itself. I don't get it, this is very confusing to me. The ANSWER for the general solution ends up to be 3x4 + 6x2y2 + 4x3 = C... Anyways:

(x2 + y2 + x)dx + (xy)dy = 0

M(x,y) = ( x2 + y2 + x ) ; N(x,y) = xy

h(x) = [ ( dM / dy ) - ( dN / dx ) ] / N ;

integrating factor = e∫h(x)dx

h(x) = (2y - y) / xy = y/xy = 1/x ; h(x) = 1/x

integrating factor = e∫(1/x)

integrating factor = eln x ; integrating factor = x


Now what do I do with it? Am I to just solve using 'grouping' or solve using 'brute force' method? Or do I take integrating factor again except using N(x,y) and then what? This is really getting me mad.
 
Physics news on Phys.org
  • #2
I'm also confused. Are you sure you're supposed to show (x^2 + y^2 + x)dx + (xy)dy = 0 is exact? But it's not. The whole point of using integrating factors is to make a non-exact equation exact.
 
  • #3
ideasrule said:
I'm also confused. Are you sure you're supposed to show (x^2 + y^2 + x)dx + (xy)dy = 0 is exact? But it's not. The whole point of using integrating factors is to make a non-exact equation exact.
That's my point :/

Should I just go ask my instructor?
 

Related to Exact Differential Equations and Integrating Factor

What is an exact differential equation?

An exact differential equation is a type of differential equation in which the partial derivatives of the solution function can be expressed as a linear combination of the variables and the solution function itself. This means that the differential equation can be solved by finding an integrating factor.

What is an integrating factor?

An integrating factor is a function that can be multiplied to both sides of an exact differential equation to make it exact. This function is typically found by taking the ratio of the coefficients of the differentials in the equation. Multiplying the integrating factor to both sides allows for easier integration and solution of the equation.

How do you find the integrating factor?

To find the integrating factor, you first need to determine if the differential equation is exact. If it is, you can then set up an equation using the coefficients of the differentials and solve for the integrating factor. This can be done by taking the ratio of the coefficients and then integrating the resulting equation.

What is the purpose of using an integrating factor?

The purpose of using an integrating factor is to make a differential equation exact, which makes it easier to solve. This is because the integrating factor allows for simpler integration of the equation, leading to a more straightforward solution.

What are some real-world applications of exact differential equations and integrating factor?

Exact differential equations and integrating factor have many applications in physics, engineering, and economics. They are used to model and solve problems related to heat transfer, fluid mechanics, electrical circuits, and population dynamics, among others. They are also essential in the study of thermodynamics and statistical mechanics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
457
  • Calculus and Beyond Homework Help
Replies
25
Views
573
  • Calculus and Beyond Homework Help
Replies
8
Views
816
  • Calculus and Beyond Homework Help
Replies
14
Views
468
  • Calculus and Beyond Homework Help
Replies
3
Views
169
  • Calculus and Beyond Homework Help
Replies
7
Views
748
  • Calculus and Beyond Homework Help
Replies
10
Views
587
  • Calculus and Beyond Homework Help
Replies
3
Views
869
  • Calculus and Beyond Homework Help
Replies
7
Views
797
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Back
Top