Separable Differential Equation (Perfect Derivative)

[scud0405]

Homework Statement

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

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.

 

[Dick]

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

 

[LCKurtz]

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

 

[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?

 

[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.

 

[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.

 

[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}

 

[scud0405]

Right. But is that the equation correctly separated?

 

[206PiruBlood]

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

 

[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.

 

[LCKurtz]

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

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

 

[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?

 

[LCKurtz]

Yes you are doing something wrong:

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

 

[scud0405]

Ahhh! Finally, it makes sense! Thanks!