I am pissed And Variation of parameters. who, what, when, where, why?

[dionysian]

Well, i really i am pissed at the moment because of this. I have been trying to figure out if there is a proof/justifaction for the method of variation of parameters and have had absolutly no luck. I have searched the internet and almost every differential equation books in my university library and none of the book give much of an explanation of the this method other than how to turn the crank and mysteriously solve equation using this method.

The source of my anger is this. it seems to me that most math books don't give explanation of why certain forumulas/method of solution actualy work, just how to use them. I don't know about everyone else but to me this is abosolutly pointless. I learn nothing at all from this.

So my questions are the following:

1) can someone point my to a resource that will explain why this method works. Please!

2) Is it just me being slightly sleep derprived/angry/bitter/undersexed/undertalented or do i have a point about math books not really focucing on why certian formulas and/or methods of solution work.

To be fair one book i found "derrick & grossman first course in differential equation" did at least try to explain the method but i don't follow it.

Here is an example of what the text says "To use this method it is necessary to know the general solution c1y1(x) + c2y2(x) of the homogenous equation y'' + a(x)y' + b(x)y = 0" ... ummm why?

and another thing the text says "Lagrange noticed that any particular solution yp of (1) must have the property that yp/y1 and yp/y2 are not constants, suggesting that we look for a particuylar solution of (1) of the form y(x) = c1(x)y1(x) + c2(x)y2(x)"... ummm i don't see why this would suggestion that particular solution.

Please physics forms your my only hope. i think i need a nap.

 

[HallsofIvy]

Well, i really i am pissed at the moment because of this. I have been trying to figure out if there is a proof/justifaction for the method of variation of parameters and have had absolutly no luck. I have searched the internet and almost every differential equation books in my university library and none of the book give much of an explanation of the this method other than how to turn the crank and mysteriously solve equation using this method.

That's strange. Every differential equations text I have ever seen has give a pretty comprehensive explanation of why "variation of paratmeters" works. What kind of "explanation" are you looking for?

The source of my anger is this. it seems to me that most math books don't give explanation of why certain forumulas/method of solution actualy work, just how to use them. I don't know about everyone else but to me this is abosolutly pointless. I learn nothing at all from this.

So my questions are the following:

1) can someone point my to a resource that will explain why this method works. Please!

2) Is it just me being slightly sleep derprived/angry/bitter/undersexed/undertalented or do i have a point about math books not really focucing on why certian formulas and/or methods of solution work.

To be fair one book i found "derrick & grossman first course in differential equation" did at least try to explain the method but i don't follow it.

Here is an example of what the text says "To use this method it is necessary to know the general solution c1y1(x) + c2y2(x) of the homogenous equation y'' + a(x)y' + b(x)y = 0" ... ummm why?

Ummm, because to BEGIN the "variation of parameters" method you have to USE the independent solutions to the homogeneous equations? And you can't do that unless you know what those solutions are?

and another thing the text says "Lagrange noticed that any particular solution yp of (1) must have the property that yp/y1 and yp/y2 are not constants, suggesting that we look for a particuylar solution of (1) of the form y(x) = c1(x)y1(x) + c2(x)y2(x)"... ummm i don't see why this would suggestion that particular solution.

If "yp/y1" and "yp/y2" are not constants, then they must be functions of x! In particular, yp/y1= f(x) so yp= f(x)y1 for some function f(x) and yp/y2= g(x) so yp= g(x)y2 for some function g(x). And, since the equation is linear, yp= f(x)y1+ g(x)y2 for some functions y1 and y2. There is nothing particularly deep about this- if an expression is not a constant, then it must be some function of x!

Please physics forms your my only hope. i think i need a nap.

Yes, take a nice long nap and then read this:

What you may be missing is the obvious fact that if yp, y1, y2 are any functions at all, there exist functions f(x) and g(x) such that yp= f(x)y1+ g(x)y2. The only difficulty is that there are an infinite number of possible f, g functions that will work! (Here's an obvious choice: take f(x)= 0, g(x)= yp/y2. Another: take f(x)= yp/y1, g(x)= 0.) That's why, after differentiating to get yp'= f' y1+ fy1'+ g' y2+ gy2', we require that f' y1+ g' y2= 0: Out of the infinite number of possible f, g functions, we search only for those that satisfy that equation just because it simplifies the equation. Now that we have yp'= f y1'+ gy2', differentiate again: yp"= f'y1'+ fy1"+ g' y2'+ g y2". If you put those values into your original equation, you we see all terms NOT involving f' and g' disappear. That's because, since you have not differentiated them, the act like constants (think carefully about that) and we already know that any constant time y1 or y2 make that quantity 0. We are left with an equation involving only f' and g' and known functions. That together with the earlier required f' y1+ g' y2= 0 gives two equations we can solve, algebraically, for f' and g'. Integrating gives us f and g- two of the infinite number of functions that would have worked!

 

[O.J.]

replyin to your last paragraph:

but if Yp = f(x) Y1 + g(x) Y2 and there are infinite f(x) and g(x) won't that mean we have infinite number of particular solutions? :S:S:S

 

[O.J.]

But this means if you have infinitely many f(x) and g(x) for Yp then you hav infinitely many particular solutions? how?

 

[HallsofIvy]

replyin to your last paragraph:

but if Yp = f(x) Y1 + g(x) Y2 and there are infinite f(x) and g(x) won't that mean we have infinite number of particular solutions? :S:S:S

But this means if you have infinitely many f(x) and g(x) for Yp then you hav infinitely many particular solutions? how?

Yes, of course there are infinitely many particular solutions! Any solution to the entire equation is a "particular solution" and there are clearly an infinite number of them- that's why you need to add the "general solution" to the homogeneous equation- to allow you to go from a single "particular solution" to any "particular solution". Look at it geometrically: a subspace of R² is a line through the origin. That corresponds to the set of solutions to a homogeneous linear differential equation. The line parallel to that but not through the origin is a "linear manifold" that is not a subspace and corresponds to the solutions to non-homogeneous differential equation. Choose any vector, $\vec{v}$ from the origin to a point on that line: there are clearly an infinite number of such vectors. Now, any vector on that line can be written as [tex]\vec{v}[/itex] plus a vector on the original line through the origin- there are clearly an infinite number of such "particular solutions", $\vec{v}$.

 

[O.J.]

I don't understand how you can havve infinitely many Yp's. Your manifold explanation is vague to me as I'm stil not familiar with such terms. (Currently taking linear algebra nad just started vector spaces). Could you clarify in another way please? And sorry for double posting.

 

[HallsofIvy]

Do you understand what a "particular solution" is? It is just a single solution to the entire equation.

Take, for example, y"- y= x. The homogeneous equation, y"- y= 0 has general solution $y= Ce^x+ De^{-x}$. If I "guess" a "particular solution" of the form y= Ax+ B, I get y'= A, y"= 0 so the equation becomes -Ax-B= x which tells me that A= -1, B= 0. The general solution to the entire equation is $y(x)= Ce^x+ De^{-x}- x$. Do you not see that $y= e^x- x$ is also a "particular solution"? And that $y= e^{-x}- x$ is another? In fact, any specific choice of C and D gives a "specific solution" to the entire equation.

 

[O.J.]

could there be another Yp from the one you foudn that is linearly independent from x?
Also, could you please explain to me how wronskians determine de-independency? not how to use them. but how they justify dependency?

 

[HallsofIvy]

could there be another Yp from the one you foudn that is linearly independent from x?

I don't know what you mean by "independent from x"? A constant function?

Also, could you please explain to me how wronskians determine de-independency? not how to use them. but how they justify dependency?

I would think any good elementary d.e. book would go through that. How much linear algebra have you had? it's hard to understand the theory behind linear des without linear algebra. A set of functions (I am going to take 2 as a simple example) is independent if and only if the only way af+ bg can equal 0 (the 0 function: 0 for all values of x) is if a and b are 0. If af+ bg= 0, then, in particular af(x₀)+ bg(x₀)= 0 for any specifice x₀. Also, differentiating the entire equation with respect to x, af'(x)+ bg'(x)= 0 for all x and so, specifically, af'(x₀)+ bg'(x₀)= 0. We now have two linear equations with numerical coefficients to solve for a and b. A system of linear equations will have a unique solution (here, obviously a= b= 0) if and only if the "determinant of coefficients" is not 0. That determinant is, in this case, precisely the Wronskian.

 

[O.J.]

I don't know what you mean by "independent from x"? A constant function?

it is related to your post where you defined a DE and found the Yp for it as x.


I would think any good elementary d.e. book would go through that. How much linear algebra have you had? it's hard to understand the theory behind linear des without linear algebra. A set of functions (I am going to take 2 as a simple example) is independent if and only if the only way af+ bg can equal 0 (the 0 function: 0 for all values of x) is if a and b are 0. If af+ bg= 0, then, in particular af(x₀)+ bg(x₀)= 0 for any specifice x₀. Also, differentiating the entire equation with respect to x, af'(x)+ bg'(x)= 0 for all x and so, specifically, af'(x₀)+ bg'(x₀)= 0. We now have two linear equations with numerical coefficients to solve for a and b. A system of linear equations will have a unique solution (here, obviously a= b= 0) if and only if the "determinant of coefficients" is not 0. That determinant is, in this case, precisely the Wronskian.

Wait a sec,
now for two functions f & g, if f is LD to g, then we can express f as f = c g => f - cg = 0 or f + ag = 0 (replaced minus c with a). But in your argument, your assuming they are LI. if they are LI f +ag can never be 0... am I making sense?

 

[slider142]

Wait a sec,
now for two functions f & g, if f is LD to g, then we can express f as f = c g => f - cg = 0 or f + ag = 0 (replaced minus c with a). But in your argument, your assuming they are LI. if they are LI f +ag can never be 0... am I making sense?

This is true unless f and g are identically 0 for some interval or x, or g corresponds to -f/a for some interval or x. This corresponds to the fact that the Wronskian corresponds to an interval, and if the Wronskian is 0, it does not imply that the functions are not linearly independent. It is only if the Wronskian is non-zero that we have a useful result. When the Wronskian is 0, you can still use the http://planetmath.org/?op=getobj&from=objects&name=GrammianDeterminant to test, but that's a little harder to compute (its the determinant of the matrix of inner products of the functions, where the inner product is the common one of integrating the product of the functions over the interval).

 

[mathwonk]

why are you angry because someone elwe does not solve your problem for you? why don't you figure it out yourself and tell others clearly? do others have more responsibility than you? are you a man or a whiny mouse?

 

[O.J.]

Im sorry to infest the thread with my questions. But now I'm even more confused about LI/D ency than ever. Could some1 please elaborate with a simple example of two functions?

 

[HallsofIvy]

"ency"?


No, I was not assuming f and g are linearly independent. The whole point of the Wronskian is to determine whether or not f and g are linearly independent. f and g will be linearly independent if the only way to have af+ bg= 0 (for all x) is to take a= b= 0. That is the same as saying that the two equations af(x₀)+ bg(x₀)= 0 and af '(x₀)+ bf '(x₀)= 0 have the unique solution. That in turn is true if and only if the determinant of coefficients, the Wronskian, is non-zero.

 

[dionysian]

why are you angry because someone elwe does not solve your problem for you? why don't you figure it out yourself and tell others clearly? do others have more responsibility than you? are you a man or a whiny mouse?

No, i do not think others have more responsibility than me, but i thought perhaps someone knew of a particular resource that has a good explanation of this. Thats all. Yes, i know i was whining, but it was 70% a joke and 30% lack of sleep and frustration. Next time i will "clearly explain" when i am kidding so you will understand.

Halls of ivy.

I did understand your explanation. Thank you. I should have mentioned that's weeks ago when you posted the response for me.

 

[mathwonk]

thank you for explaining when you are whining out of 70% humorousness. this helps me greatly.

 

[notpigfly]

That's strange. Every differential equations text I have ever seen has give a pretty comprehensive explanation of why "variation of paratmeters" works. What kind of "explanation" are you looking for?


Ummm, because to BEGIN the "variation of parameters" method you have to USE the independent solutions to the homogeneous equations? And you can't do that unless you know what those solutions are?


If "yp/y1" and "yp/y2" are not constants, then they must be functions of x! In particular, yp/y1= f(x) so yp= f(x)y1 for some function f(x) and yp/y2= g(x) so yp= g(x)y2 for some function g(x). And, since the equation is linear, yp= f(x)y1+ g(x)y2 for some functions y1 and y2. There is nothing particularly deep about this- if an expression is not a constant, then it must be some function of x!


Yes, take a nice long nap and then read this:

What you may be missing is the obvious fact that if yp, y1, y2 are any functions at all, there exist functions f(x) and g(x) such that yp= f(x)y1+ g(x)y2. The only difficulty is that there are an infinite number of possible f, g functions that will work! (Here's an obvious choice: take f(x)= 0, g(x)= yp/y2. Another: take f(x)= yp/y1, g(x)= 0.) That's why, after differentiating to get yp'= f' y1+ fy1'+ g' y2+ gy2', we require that f' y1+ g' y2= 0: Out of the infinite number of possible f, g functions, we search only for those that satisfy that equation just because it simplifies the equation. Now that we have yp'= f y1'+ gy2', differentiate again: yp"= f'y1'+ fy1"+ g' y2'+ g y2". If you put those values into your original equation, you we see all terms NOT involving f' and g' disappear. That's because, since you have not differentiated them, the act like constants (think carefully about that) and we already know that any constant time y1 or y2 make that quantity 0. We are left with an equation involving only f' and g' and known functions. That together with the earlier required f' y1+ g' y2= 0 gives two equations we can solve, algebraically, for f' and g'. Integrating gives us f and g- two of the infinite number of functions that would have worked!

Dear HallsofIvy, can u recommand some nice ODE books which have a nice intuition about some methods like varation of parameters, undetermined coefficient also including theoritial depth?

THX :D