# Separable Differential Equation (Perfect Derivative)

1. Oct 7, 2009

### scud0405

1. The problem statement, all variables and given/known data

$$-y + xy' = 0$$ and y(2)=5

3. The attempt at a solution

This first part trips me up. I am supposed to find the perfect derivative, which is $$(xy') = 0$$ ? Is this legal, or does the -y not allow for that?

If that is correct, then I know that I integrate that, which gives me $$xy = c$$

Then, to solve for c, I just plug in 2 and 5.

2. Oct 7, 2009

### Dick

No, it's not exact. And your solution assuming that is wrong. But it is separable. Try that.

3. Oct 7, 2009

### LCKurtz

It's also linear. Have you studied integrating factors?

4. Oct 8, 2009

### scud0405

Oh, I understand! Just multiply by e^-x, so the problem looks like this:

$$-ye^-^x + xe^-^xy' = 0$$

Therefore:

$$xe^-^xy = c$$

Correct?

I'm having trouble with the next problem also. The next problem is similar, but with a $$x^3$$ added to it. I can't think of any other way to do it other than by multiplying everything by e^-x and doing integration by parts to $$x^3e^-^x$$.Any tips for an easier way to do it?

Last edited: Oct 8, 2009
5. Oct 8, 2009

### LCKurtz

No, your solution isn't correct. Your original equation

$$xy' - y = 0$$

is not in the correct form to find the integrating factor. The coefficient of y' needs to be 1 so you need to divide by x before calculating the integrating factor. Also, as a side note, linear differential equations are typically written with the highest derivatives first much like polynomials, not that it matters that much.

6. Oct 8, 2009

### scud0405

Oh.. you're right.

It's not separable though, is it? When I try and separate it I cannot get the Ys on one side and the Xs on one side. Unless $$\frac{dy}{y} = \frac{dx}{x}$$ is the equation correctly separated.

7. Oct 8, 2009

### 206PiruBlood

remember that is $$\frac{1}{x}dx=\frac{1}{y}dy$$ You can integrate this.

As for the other method, do you know how to find an integrating factor for second order linear equations? It's generally not just multiply through by $$e^{-x}$$

Last edited: Oct 8, 2009
8. Oct 8, 2009

### scud0405

Right. But is that the equation correctly separated?

9. Oct 8, 2009

### 206PiruBlood

Yes but I believe the problem is asking you to find an integrating factor.

10. Oct 8, 2009

### scud0405

Okay, so if I divide by X, I get $$y' - \frac{y}{x} = 0$$

So the integrating factor would be -x? That can't be right because that just gives me what I started out with.

11. Oct 8, 2009

### LCKurtz

For y' + P(x)y = Q(x) the integrating factor is

$$R(x) = e^{\int P(x) dx}$$

12. Oct 8, 2009

### scud0405

$$e^\int^-^1^/^x$$ = $$e^-^l^n^x$$ = -x. So this means I would multiply the whole problem by -x which gives me what I start out with.

Or am I doing something wrong?

13. Oct 8, 2009

### LCKurtz

Yes you are doing something wrong:

$$e^{- \ln(x)} = e^{\ln(x^{-1})}= x^{-1}$$

14. Oct 8, 2009

### scud0405

Ahhh! Finally, it makes sense! Thanks!