# Another First Ord. Diff. Equation

1. Jan 18, 2004

### wubie

Hello,

Once again I am having trouble verifying my answer.

Here is the question:

Solve the given first-order linear equation and verify that your solution indeed satisfies the equation:

dy/dx = y/x + 1/x^2, x>0

Now I get an answer of

y = -1/2 * 1/x + xC

But when I sub. back in the the equation

dy/dx = y/x + 1/x^2

It doesn't verify. I keep on trying it over and over again, but I keep on getting the same thing. What am I doing wrong?

Any help is appreciated. Thankyou.

2. Jan 18, 2004

### jamesrc

I get

$$y = -Cx - \frac{1}{2x}$$

I didn't verify it, so that may not be it, but maybe it is.

EDIT: Just verified it, that's should be the answer; looks like you just missed a sign somewhere.

3. Jan 18, 2004

### wubie

How did you get

-Cx?

I get this:

1/x * y = integral of 1/x^3

which becomes

1/x * y = -1/2 * 1/x^2 + C

then

y = -1/2 * 1/x + Cx

How did you get -Cx?

Some more nfo.

When I try to find the integrating factor I know that

f(x) = - 1/x

Integrating that I get

- ln x

which is the same as

ln x^-1

When this is to the power of e

e^ln x^-1 = 1/x

This is correct right?

So my integrating factor should be 1/x right?

Last edited by a moderator: Jan 18, 2004
4. Jan 18, 2004

### jamesrc

First, I found the integrating factor to get an exact equation:

- rearrange $$\frac{dy}{dx} = \frac{y}{x} + \frac{1}{x^2}$$ to get

$$M(x,y)dx + N(x,y)dy = 0$$
$$(yx + 1)dx + (-x^2)dy = 0$$

find $$\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{x-(-2x)}{-x^2} = -\frac 3 x$$

$$u(x) = e^{-\int{\frac{3}{x}dx}} = x^{-3}$$

Now multiply the factor (the equation is now exact, but I'll skip that step (you can check if you want)):

$$\frac{yx+1}{x^3}dx + (-\frac 1 x)dy = 0$$

So:

$$F(x,y) = C = \int{N(x,y)dy} + \Phi(x) = -\frac y x + \Phi(x)$$

$$\frac{\partial F}{\partial x} = \frac{y}{x^2} + \Phi'(x) = M(x,y) = \frac{yx + 1}{x^3}$$

Solve for &Phi;(x) (using algebra and integration) and then find F(x,y):

$$F(x,y) = -\frac y x - \frac{1}{2x^2} = C$$

Solving for y:

$$y = -Cx - \frac{1}{2x}$$

which can be verified as the correct answer. Does that help?

Last edited: Jan 18, 2004
5. Jan 18, 2004

### wubie

I'm not too familiar with the first part. If that is the way you did it then that is where I got into trouble. I am not too familiar with that notation either.

Typically we try to get the equation into a form of

y' + F(x)y = Q(x)

and then try to find the integrating factor as such

I(x) = e^integral of F(x) dx

So I would then take the integral of F(x).

I will have to look over the first part some more and try to figure it out. I might need some help in the first part. If that is not too much trouble.

Thanks for your help so far.

6. Jan 18, 2004

### wubie

This is what we were shown to do in the text and lecture:

Put into the form y' + P(x)y = Q(x)

So that is what I did:

y' - 1/x * y = 1/x^2.

From that I have P(x) = -1/x.

Now I find the integrating factor I(x) where

I(x) = e^integral of P(x) dx

So then

e^integral of -1/x dx

becomes

e^lnx^-1

which is the same as

x^-1 = 1/x

That is how we are supposed to find the integrating factor. I haven't your procedure before. If you say your answer is correct I would have to believe you since my way is not checking out.

Can you see any similarity to the way you are doing it and the way I am doing it? And if so, can you show me those similarities? If it isn't too much trouble.

7. Jan 19, 2004

### jamesrc

Ugh. We're both right. Well, you're more right because I didn't realize you were right.

Anyway, the sign on an arbitrary constant C clearly doesn't matter, so

$$y = C_1x- \frac{1}{2x} = -C_2x-\frac{1}{2x}$$

same difference.

Doublecheck your verification step; it should work out:
$$y' = C + \frac{1}{2x^2}$$
$$\frac y x = C - \frac{1}{2x^2}$$
$$y' - \frac y x + \frac{1}{x^2} = (C + \frac{1}{2x^2}) - (C - \frac{1}{2x^2}) - \frac{1}{x^2} = 0$$

So that works out.

Sorry about any foreign notation/formulation I used. I'm a little rusty at this myself and that's just the way I remembered it. Rest assured that both of our formulations accomplish the same task.

8. Jan 19, 2004

### wubie

Hi James,

First of all, I really appreciate your time and effort. Maybe I have just been looking at this too long.

When I did what you said and did the verification yet again, it worked out. I must have had some sort of brain malfunction or something.

Thanks again James.

Have a good night.