Solving Linear Differential Equations: Explained

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In summary: G(x) = xSo in summary, the general form of a linear differential equation is y' + P(x)y = G(x). In the given equation, P(x) is equal to -1 and G(x) is equal to x.
  • #1
suspenc3
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Im confused concerning linear diff eqns. I can solve some of them but I don't understand something in the following form.

Say [tex]dy/dx=x+y[/tex]

what is [tex]P(x)[/tex] going to be?

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


Thanks.
 
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  • #2
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
suspenc3 said:
Im confused concerning linear diff eqns. I can solve some of them but I don't understand something in the following form.

Say [tex]dy/dx=x+y[/tex]

what is [tex]P(x)[/tex] going to be?

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


Thanks.
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
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 [tex]e^{\int 2x dx}[/tex]
 
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  • #5
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
[tex]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
[itex]e^{\int 2dx}= e^{2x}[/itex].
[tex](e^{2x}y)'= e^{2x}y'+ 2e^{2x}y[/tex]

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
Ok, I get it now, i was just confused with the P(x)y and there being no "x"
 
  • #7
suspenc3 said:
Im confused concerning linear diff eqns. I can solve some of them but I don't understand something in the following form.

Say [tex]dy/dx=x+y[/tex]

what is [tex]P(x)[/tex] going to be?

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


Thanks.

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

[tex]y' + P(x)y = G(x)[/tex]

So now just arrange your equation to look like that:

[tex]y' - y = x [/tex]

or

[tex]y' + (-1) y = x [/tex]

From this you can see that:

[tex]P(x) = -1[/tex]
 

1. What is a linear differential equation?

A linear differential equation is a mathematical equation that represents the relationship between a variable and its derivatives. It is called "linear" because the variable and its derivatives are only raised to the first power, and there are no products or compositions involving them.

2. How do you solve a linear differential equation?

To solve a linear differential equation, first separate the variables (the function and its derivatives) on opposite sides of the equation. Then, integrate both sides and add a constant of integration. Finally, solve for the constant by using initial conditions or boundary conditions given in the problem.

3. What is the difference between an ordinary and a partial differential equation?

An ordinary differential equation (ODE) involves a single independent variable and its derivatives. A partial differential equation (PDE) involves multiple independent variables and their partial derivatives. In other words, an ODE deals with functions of one variable, while a PDE deals with functions of multiple variables.

4. What are some real-life applications of solving linear differential equations?

Linear differential equations are used in many fields of science and engineering, including physics, chemistry, biology, and economics. They are used to model and predict the behavior of systems such as population growth, chemical reactions, and circuits.

5. Are there any specific techniques for solving linear differential equations?

Yes, there are various techniques for solving linear differential equations, including separating the variables, using the integrating factor, and using the method of undetermined coefficients. The specific technique used depends on the form of the equation and the given initial/boundary conditions.

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