Not understanding integrating factorsplease clarify/explain

In summary: So i can do integ by parts.In summary, the conversation discusses the steps necessary to solve a specific differential equation by finding an integrating factor, including the use of exactness and integration by parts. The conversation also clarifies the meaning of a homogeneous equation and the role of mu(x) in the solution process.
  • #1
darryw
127
0

Homework Statement


Specifically what i don't understand is a situation where you can't get the equation in the standard form of y' + p(x)y = q(x)

the integrating factor is e^integ p(x), but what if y is part of a sin or cos, as in this equation:
(x+2)sin y dx + xcos y dy = 0

if i rearrange so that it is: y' + (x+2)sin y / x cos y = 0, it still isn't in standard form.

So at that point how am i supposed to proceed? What step do i take to get the integrating factor?

So, if equation isn't separable, and isn't exact, and can't be put into standard form, how do you solve it?
thanks for any help

Homework Equations





The Attempt at a Solution


 
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  • #2
darryw said:

Homework Statement


Specifically what i don't understand is a situation where you can't get the equation in the standard form of y' + p(x)y = q(x)

the integrating factor is e^integ p(x), but what if y is part of a sin or cos, as in this equation:
(x+2)sin y dx + xcos y dy = 0

if i rearrange so that it is: y' + (x+2)sin y / x cos y = 0, it still isn't in standard form.

So at that point how am i supposed to proceed? What step do i take to get the integrating factor?

So, if equation isn't separable, and isn't exact, and can't be put into standard form, how do you solve it?
thanks for any help

You are confusing two different techniques. A differential equation of the form y' + p(x)y = q(x) is a linear, first-order, nonhomogeneous equation. It's linear because y and its derivative occur to the first power. There's a technique that involves finding an integrating factor that can be used to solve this type of equation.

The equation (x + 2)sin y dx + xcosy dy = 0 can be written as y' + (1 + 2/x)tany = 0, so this is also first-order, but it's nonlinear because y appears as part of a trig function. The standard approach for equations written as (x + 2)sin y dx + xcosy dy = 0 is to determine whether the equation is exact. If so, then this equation represents the total derivative of some as-yet-undetermined function F(x, y). As you have already found out, this equation is not exact, so that approach isn't applicable.

What other techniques do you know?
 
  • #3
Well the problem asks me to solve by finding an integrating factor, mu(x), so that's why I am stuck on getting an integrating factor. I also know about 1st order homogeneous equations but i don't think that applies here because I can't manipulate it so that y/x. ( v = y/x)
thank you
 
  • #4
multiply both sides by mu(x)??
 
  • #5
darryw said:
multiply both sides by mu(x)??
Is this the technique where you assume there is an integrating factor mu(x) and work it into the equation so as to make the equation exact? It's been awhile for me since I thought about this.
 
  • #6
darryw said:
Well the problem asks me to solve by finding an integrating factor, mu(x), so that's why I am stuck on getting an integrating factor. I also know about 1st order homogeneous equations but i don't think that applies here because I can't manipulate it so that y/x. ( v = y/x)
thank you
Unfortunately two different mathematicians came up with categorizations for d.e.s and both settled on the same term for completely different things - homogeneous. The meaning that seems to me to be used most often is when you have a differential equation of the form f(y, y', y'', ...) = 0. E.g., the one in another thread you started - y'' + 3y' + 2y = 0.

The other meaning - that you can write the diff. equation as f(y/x) or something seems much more obscure to me, and doesn't seem to appear as much.
 
  • #7
Mark could you please tell me what you think best technique to use on this? ,--the thing is, i could never really grasp the professors notes so I am learning most of this off internet and I have exam tomorrow, so what do you personally think is best method (easiest) to solve it? --if you won't show me the steps can you tell me most recommended method so i can learn it, then apply it here? ..thanks for any more help.
 
  • #8
The specific question asks:
Solve (x + 2) sin y dx + x cos y dy = 0 by finding an integrating factor, mu(x).
(from homework probs)
thanks
 
  • #9
Your equation doesn't appear to be homogeneous in the sense of being y' = f(y/x), but it is solvable by applying the right integrating factor [itex]\mu[/itex], where [itex]\mu[/itex] can be a function of x ([itex]\mu[/itex] = [itex]\mu[/itex](x)) or of y ([itex]\mu[/itex] = [itex]\mu[/itex](y)).


Let's start with [itex]\mu[/itex](x).
If we multiply both sides of your original equation by [itex]\mu[/itex](x), we get
[[itex]\mu[/itex](x) * (x + 2) * sin y] dx + [[itex]\mu[/itex](x) * x * cosy] dy = 0

The expression in the first pair of brackets is M, and the expression in the second pair of brackets in N.

For exactness, we must have My = Nx. Can you continue from here?

It's possible in certain cases to get a differential equation for [itex]\mu[/itex](x) that involves both x and y, which wouldn't be solvable. In that case, we take exactly the same approach, but with [itex]\mu[/itex] as a function of y.
 
  • #10
thanks.

[mu(x) * (x + 2) * sin y] dx + [mu(x) * x * cosy] dy = 0
...M........N

M_y(x,y) = cos y

N_x(x,y) = cos y

so now it is exact.

so now i integrate M wrt x, and then take derivative of that result and equate it to N and solve for h'(y). Then integrate that to get h(y) and write as final equation.

But how am i supposed to view mu(x) ?? Is it a constant?
I mean, when integrating wrt x the M term:

mu(x)(x+2)siny there are three terms, so how does integ by parts apply or even u-sub apply?

thanks

EDIT: ok if I am integ wrt x there are two terms multiplied, mu(x) * (x+2) and the sin y is treated as constant.
but I am still confused how to treat mu(x). thanks
 
Last edited:
  • #11
darryw said:
thanks.

[mu(x) * (x + 2) * sin y] dx + [mu(x) * x * cosy] dy = 0
...M........N

M_y(x,y) = cos y

N_x(x,y) = cos y

so now it is exact.
Not so fast. In differentiation with respect to y, x is seen as a constant, so My = mu(x) * (x + 2) * cos y. For Nx, you have to use the product rule.
darryw said:
so now i integrate M wrt x, and then take derivative of that result and equate it to N and solve for h'(y). Then integrate that to get h(y) and write as final equation.

But how am i supposed to view mu(x) ?? Is it a constant?
I mean, when integrating wrt x the M term:

mu(x)(x+2)siny there are three terms, so how does integ by parts apply or even u-sub apply?

thanks

EDIT: ok if I am integ wrt x there are two terms multiplied, mu(x) * (x+2) and the sin y is treated as constant.
but I am still confused how to treat mu(x). thanks
 
  • #12
My = mu(x) * (x + 2) * cos y

Nx = (mu'(x)*x + mu(x) ) cos y

So not exact afterall ?
 
  • #13
It depends on what mu(x) is. What does it take for the d.e. to be exact? If it's exact, then My = Nx, right?

Can you solve for mu(x) so that the equation is exact?
 

1. What is an integrating factor?

An integrating factor is a function used in solving a first-order differential equation that is not exact. It is multiplied to both sides of the equation in order to make it exact and easier to integrate.

2. Why is it important to use integrating factors?

Integrating factors are important because they allow us to solve a wider range of first-order differential equations that would otherwise be difficult or impossible to solve. They also simplify the integration process and make it more manageable.

3. How do I determine which integrating factor to use?

The integrating factor can be determined by examining the coefficients of the differential equation and using a formula or method that is specific to the type of equation. For example, for linear equations, the integrating factor is typically e raised to the integral of the coefficient of the independent variable.

4. Can I solve a differential equation without using an integrating factor?

In some cases, it is possible to solve a differential equation without using an integrating factor. However, this may require more advanced mathematical techniques and may not always be possible. Using an integrating factor is often the most efficient and straightforward method.

5. How do I know if I have solved the differential equation correctly using an integrating factor?

You can check your solution by plugging it back into the original differential equation. If it satisfies the equation, then you have solved it correctly. Additionally, you can use a graphing calculator or software to plot the solution and compare it to the graph of the original differential equation to ensure they match.

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