ODE, Integrating Factor method

In summary: Simplifying gives you y(x)=(ln(x)+2)/x.In summary, the conversation is about solving a differential equation using the integrating factor method. The steps involved include dividing through by x^2, computing the integrating factor (x), and using the product rule to find d/dx(x*y(x)). The final step is to plug in a given value and solve for C, which is then substituted back into the equation to find the final solution.
  • #1
tony873004
Science Advisor
Gold Member
1,752
143
I'm trying to follow this example from class notes.

[tex]x^2 \frac{{dy}}{{dx}} + xy = 1,\,\,y\left( 1 \right) = 2[/tex]

Divide through by x^2 to set it up in a form for using the integrating factor method.

[tex]
\frac{{dy}}{{dx}} + \frac{y}{x} = - \frac{1}{{x^2 }}

[/tex]
I'm not sure where the minus sign came from on the right side. I just copied the problem from the board.

Seperate the functions of x to make it in I.F. form
[tex]
\frac{{dy}}{{dx}} + y\left( x \right)\frac{1}{x} = - \frac{1}{{x^2 }}
[/tex]

Compute the Integrating Factor with from the term that doesn't contain a y. Is this correct? Is this why 1/x was chosen to integrate and raise e to the power of it?

[tex]
I.F.\,\,e^{\int_{}^{} {\frac{1}{x}dx} } = e^{\ln x} = x
[/tex]

The next step in the notes is
[tex]
x\frac{{dy}}{{dx}} + y = \frac{1}{x}
[/tex]
Is this simply from multiplying the Integrating Factor, in this case, x, by all 3 terms in the 2nd equation? If so, what happened to my minus sign (or did it never exist and I inadvertantly put it in by mistake)?

The problem is longer than this, but I want to understand it up to this point for the moment. Thanks!
 
Physics news on Phys.org
  • #2
Yes. You have the right idea. I don't know why the minus sign is appearing and disappearing on the right side. Somebody is just being careless.
 
  • #3
Thanks. That careless person may be me. I don't always copy flawless notes :)

So I left off at [tex]x\frac{{dy}}{{dx}} + y = \frac{1}{x}[/tex]

The next step (assuming I copied it correctly) is
[tex]
\left( {xy} \right)^\prime = \frac{1}{x}
[/tex]

The right side remained the same. So how is it that [tex]
x\frac{{dy}}{{dx}} + y = \left( {xy} \right)^\prime
[/tex]
?
This is the same as saying [tex]
xy' + y = \left( {xy} \right)^\prime

[/tex], isn't it?
 
  • #4
tony873004 said:
Thanks. That careless person may be me. I don't always copy flawless notes :)

So I left off at [tex]x\frac{{dy}}{{dx}} + y = \frac{1}{x}[/tex]

The next step (assuming I copied it correctly) is
[tex]
\left( {xy} \right)^\prime = \frac{1}{x}
[/tex]

The right side remained the same. So how is it that [tex]
x\frac{{dy}}{{dx}} + y = \left( {xy} \right)^\prime
[/tex]
?
This is the same as saying [tex]
xy' + y = \left( {xy} \right)^\prime

[/tex], isn't it?

Sure. Just use the product rule to figure out d/dx(x*y(x)).
 
  • #5
still a bit confused. Product rule makes sense as a check. I can go from (xy)' = x'y+xy'=y+xy'. When we learned the product rule, we learned to take a product and turn it into two terms. But what's the method for doing the product rule in reverse? How do I start with two terms and work them into a product?
 
  • #6
That's the magic of the integrating factor. If the integrating factor is M(x) and the left side of the ode is y'+a(x)*y(x), you know that the derivative (M(x)*y(x))' will come out to be M(x)y'(x)+M(x)*a(x)*y(x). That's why it's called an 'integrating factor'.
 
  • #7
Thanks. I'll take me a little while to grasp that. There's one final question in this problem. After integrating both sides and ending up with
[tex]
y = \frac{{\ln x + C}}{x}

[/tex]

I plug in 1 for x and set the equation to 2, as per the instructions.
[tex]
y\left( 1 \right) = \frac{{\ln \left( 1 \right) + C}}{1} = 2
[/tex]

The next step (which is the answer) from the class notes is simply
[tex]
y = \frac{{\ln x + 2}}{x}
[/tex]

How did this last step come about? I hate to say that the 1's magically turned back into x's, and the answer replaced C.

gotta run out the door now, so if I don't says 'thanks' for a few hours, I still appreciate your explanations.
 
  • #8
There's nothing magic about it. You solved y(1)=(ln(1)+C)/1=2 for C. Then you plugged that value of C back into y(x)=(ln(x)+C)/x.
 

Related to ODE, Integrating Factor method

1. What is an Ordinary Differential Equation (ODE)?

An Ordinary Differential Equation (ODE) is a mathematical equation that describes how a variable changes over time. It involves one or more independent variables and their derivatives with respect to the independent variables.

2. What is the Integrating Factor method?

The Integrating Factor method is a technique used to solve first-order linear differential equations. It involves multiplying the entire equation by an integrating factor, which is a function of the independent variable.

3. How does the Integrating Factor method work?

The Integrating Factor method works by multiplying both sides of the differential equation by an integrating factor, which is chosen to make the equation easier to solve. This factor is usually a function of the independent variable, and its purpose is to "integrate" the equation and make it solvable.

4. When is the Integrating Factor method used?

The Integrating Factor method is used to solve first-order linear differential equations. These are equations where the dependent variable and its derivative are linearly related to the independent variable. It is also used when other methods, such as separation of variables, are not applicable.

5. What are some advantages of using the Integrating Factor method?

The Integrating Factor method is a powerful technique that can be used to solve a wide range of first-order linear differential equations. It is relatively easy to use and can often be applied to problems that cannot be solved using other methods. It also allows for the solution of differential equations with initial conditions, making it more versatile than some other techniques.

Similar threads

  • Calculus and Beyond Homework Help
Replies
19
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
234
  • Calculus and Beyond Homework Help
Replies
8
Views
889
  • Calculus and Beyond Homework Help
Replies
3
Views
387
  • Calculus and Beyond Homework Help
Replies
10
Views
689
  • Calculus and Beyond Homework Help
Replies
2
Views
538
  • Calculus and Beyond Homework Help
Replies
3
Views
504
  • Calculus and Beyond Homework Help
Replies
5
Views
802
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
688
Back
Top