ODE, Integrating Factor method

[tony873004]

I'm trying to follow this example from class notes.

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

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

<br /> \frac{{dy}}{{dx}} + \frac{y}{x} = - \frac{1}{{x^2 }}<br /> <br />
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
<br /> \frac{{dy}}{{dx}} + y\left( x \right)\frac{1}{x} = - \frac{1}{{x^2 }}<br />

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?

<br /> I.F.\,\,e^{\int_{}^{} {\frac{1}{x}dx} } = e^{\ln x} = x<br />

The next step in the notes is
<br /> x\frac{{dy}}{{dx}} + y = \frac{1}{x}<br />
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!

 

[Dick]

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.

 

[tony873004]

Thanks. That careless person may be me. I don't always copy flawless notes :)

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

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

The right side remained the same. So how is it that <br /> x\frac{{dy}}{{dx}} + y = \left( {xy} \right)^\prime <br />
?
This is the same as saying <br /> xy&#039; + y = \left( {xy} \right)^\prime <br /> <br />, isn't it?

 

[Dick]

Thanks. That careless person may be me. I don't always copy flawless notes :)

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

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

The right side remained the same. So how is it that <br /> x\frac{{dy}}{{dx}} + y = \left( {xy} \right)^\prime <br />
?
This is the same as saying <br /> xy&#039; + y = \left( {xy} \right)^\prime <br /> <br />, isn't it?

Sure. Just use the product rule to figure out d/dx(x*y(x)).

 

[tony873004]

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?

 

[Dick]

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'.

 

[tony873004]

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
<br /> y = \frac{{\ln x + C}}{x}<br /> <br />

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

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

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.

 

[Dick]

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.