# Homework Help: Linear diff eqns

1. Aug 27, 2006

### suspenc3

Im confused concerning linear diff eqns. I can solve some of them but I dont understand something in the following form.

Say $$dy/dx=x+y$$

what is $$P(x)$$ going to be?

Not only for this, but a general explanation would help.

Thanks.

2. Aug 27, 2006

### d_leet

I'm not sure what P(x) is the "coefficient" of in the general form you are used to so I'll try to answer this the best I can.

Let's look at

j(x)y' + k(x)y = l(x)
using j, k and l to avoid confusion with p,q, and r.

Then for the equation you have
y' = x + y
so
y' - y = x
then in this case
j(x) = 1
k(x) = -1
l(x) = 1

Does this help at all?

3. Aug 28, 2006

### HallsofIvy

You have to understand that just saying "P(x)" doesn't tell us what you are looking for. There is no standard understanding of P(x)- except that perhaps it is some polynomial in x. Because you give a linear, first order, differential equation, y'- y= x, here are some ways I might interpret it:

1) P(x) is the "characteristic polynomial".
If we try a solution of the form y(x)= erx, to the "homogeneous part" of the equation, y'- y= 0, then y'(x)= rerx so y'- y= rerx- erx= 0. Since erx is never 0, we can divide the equation by it and get the "characteristic equation" r- 1= 0 which must be satisfied by r in order that erx be a solution to the equation. The "characteristic polynomial" then is P(r)= r-1 or P(x)= x-1. That has the advantage that it is a polynomial as implied by the notation "P(x)" although it is strange that you should write it as a function of x, the independent variable in the differential equation.

1) P(x) is the "integrating factor".
Every first order differential equation has an integrating factor although the may be impossible to find. For a linear first order differential equation, it is particularly easy to find- there is a simple formula. I will, instead of using the formula, work it out from the definition.
We can write the differential equation as y'- y= x. An "integrating factor" for the equation is a function (not necessarily a polynomial) P(x) such that multiplying the equation by P(x) makes the left side a single derivative. That is, so that P(x)y'- P(x)y= d(P(x)y)/dx. Using the chain rule to expand the right side of that, we must have P(x)y'- P(x)y= P(x)y'+ P'(x)y. We can subtract P(x)y' from both sides and divide through by y. That gives P'(x)= -P(x) which is satisfied by P(x)= e-x. That is, multiplying the equation y'- y= x by e-x we have
e-xy'- e-xy= (e-xy)'= xe-x. We can integrate the left side immediately to get e-xy and integrate the right side by parts.

3. P(x) is simply the "non-homogeneous" part of the differential equation.
Writing y'= x+ y as y'- y= x, so that everything involving y is is on the left side and all terms not involving y are on the right, the "homogenous part" of the equation is y' -y and the "non-homogenous" part is P(x)= x (in general this is not necessarily a polynomial).

Unless you give us more information about what you mean by "P(x)", we can't be sure which, if any, of these is meant.

4. Aug 28, 2006

### suspenc3

Yes that explains it quite well. Basically what I meant was in the form
dy/dx + P(x)y = g(x), I would understand if P(x)y was say...4xy, but if it is just 2y, then would I(x) be $$e^{\int 2x dx}$$

Last edited: Aug 28, 2006
5. Aug 28, 2006

### HallsofIvy

If you are really talking about the P(x) in the form dy/dx+ P(x)y= g(x) then for your equation, dy/dx= x+ y, since it is of the form dy/dx- y= x,
P(x)= -1 and g(x)= x. But then you refer to I(x) which, I take it, is the integrating factor.

Yes, the integrating factor for a linear first order differential equation of the form dy/dx+ P(x)y= g(x) is
$$I(x)= e^{\int P(x)dx}[/itex]. You can get that, as I said before, by recognizing that you want I(x)dy/dx+ I(x)P(x)y= d(I(x)y)/dx. Differentiating the right side with the product law and cancelling as before, dI/dx= P(x)I, a separable differential equation. Then dI/I= P(x)dx giving the exponential formula. No, if P(x)y is "just 2y" then P(x)= 2 not 2x. The integrating factor is $e^{\int 2dx}= e^{2x}$. [tex](e^{2x}y)'= e^{2x}y'+ 2e^{2x}y$$

In your original example, dy/dx= x+ y, dy/dx- y= x so "P(x)" is -1. The integrating factor is I(x)= e-x.

6. Aug 28, 2006

### suspenc3

Ok, I get it now, i was just confused with the P(x)y and there being no "x"

7. Jul 24, 2007

### EugP

If i'm not mistaken, the general equation is this:

$$y' + P(x)y = G(x)$$

So now just arrange your equation to look like that:

$$y' - y = x$$

or

$$y' + (-1) y = x$$

From this you can see that:

$$P(x) = -1$$