What is Variation of Parameters ?

[alpha754293]

What is "Variation of Parameters"?

Homework Statement

None. General.

Homework Equations

I don't know. :( ?

The Attempt at a Solution

?

I am taking a class right now on engineering analysis (which I am finding it to be more like partial differential equations mixed with boundary value problems on steroids) and the way that my professor is explaining stuff to us doesn't make a lot of sense to me and I've tried asking him and it doesn't really seem to be helping.

I've also tried looking it up on the internet as well, and they all pretty much say the same thing, which doesn't really help me understand it either.

So...what IS "variation of parameter"?

Are they really good examples that show how it is used and how to solve problems with it?

We don't have TA's or anything like that at my school, and because it's considered to be a graduate level class, we also don't have any tutors for it either. :(

We were told that we had to use this method in the one of the programs that was posted on here: https://www.physicsforums.com/showthread.php?t=265432&highlight=cauchy-euler

And I didn't understand what it meant by "using variation of parameters' method.

So any help that can explain what my prof is talking about would be GREATLY appreciated!

 

[rock.freak667]

"Wikipedia"[/URL] has the basic proof I was going to type out.

It is just to basically apply two formulas and find y.

 

[alpha754293]

In the example problem that I have linked above, how would I go about using it to solve it?

(I understand that it might be too much to ask for a full solution, but it would definitely help for me to try and understand it.)

I read it on Wikipedia and a few other places and it hasn't doesn't help me understand it any better. :(

 

[HallsofIvy]

The problem given is "$x^2y"+ axy+ by= 0$". We are given that $y_1= x^m$ is a solution and are asked to use "variation of parameters" to find another solution.

Try a solution of the form $y(x)= x^mu(x)$ for some unknown function u(x). Then $y'= mx^{m-1}u+ x^mu'$ and $y"= m(m-1)x^{m-2}u+ 2mx^{m-1}u'+ x^mu"$. Putting those into the equation gives
$$m(m-1)x^mu+ 2mx^{m+1}u'+ x^{m+2}u"+ amx^mu+ ax^{m+1}u'+ bx^mu$$

[tex]= x^m\left[m(m-1)+am+ b\right]u+ x^{m+1}(xu"+ (2m+ a)u')= 0[/itex]

Because were are told that $x^m$ satisfies the equation, we must have $m(m-1)+ am+ b= 0$. From that, we can divide both sides of the equation by $x^m$ and get
[tex]xu"+ (2m+ a)u'= 0[/itex]

Let v= u' and that becomes the easy, separable, first order equation xv'+ (2m+a)v= 0 or
dv/v= (2m+a)dx/x.

 

[alpha754293]

The problem given is "$x^2y"+ axy+ by= 0$". We are given that $y_1= x^m$ is a solution and are asked to use "variation of parameters" to find another solution.

Try a solution of the form $y(x)= x^mu(x)$ for some unknown function u(x). Then $y'= mx^{m-1}u+ x^mu'$ and $y"= m(m-1)x^{m-2}u+ 2mx^{m-1}u'+ x^mu"$. Putting those into the equation gives
$$m(m-1)x^mu+ 2mx^{m+1}u'+ x^{m+2}u"+ amx^mu+ ax^{m+1}u'+ bx^mu$$

[tex]= x^m\left[m(m-1)+am+ b\right]u+ x^{m+1}(xu"+ (2m+ a)u')= 0[/itex]

Because were are told that $x^m$ satisfies the equation, we must have $m(m-1)+ am+ b= 0$. From that, we can divide both sides of the equation by $x^m$ and get
[tex]xu"+ (2m+ a)u'= 0[/itex]

Let v= u' and that becomes the easy, separable, first order equation xv'+ (2m+a)v= 0 or
dv/v= (2m+a)dx/x.

okay...so if I understand this correctly -- variation of parameters uses one equation and one known solution to find a second equation.

and then using the first equation and the new equation that we just found, we apply that to the original problem in order to try to solve it?

Do I have that right?

(I'm trying to read LaTeX, but it doesn't render properly in my brain. (I'd wished that people would just write the equations in like Word or something and then save it as a PNG or GIF or something), but I suppose that would be to difficult, and not everybody uses word.)

 

[rock.freak667]

I think HallsofIvy did reduction of order where you know one solution,y₁, and use the fact that the other solution is y=vy₁



But this is the basic method of how to solve a DE using the method of variation of paramters.

Solve: y''+y= sinx/cosx

using the auxiliary equation (since the DE has constant coeffcients), we get

m²+1=0 => $0 \pm i$ ( this implies a solution of Acosx+Bsinx)

So we let y₁=sinx and y₂=cosx

We now find the wronskian of y1 and y2, W(y1,y2) which is given by
|sinx cosx|
|cosx -sinx|

thus W(y1,y2)= -sin²x-cos²x=-1.


and the answer we want is y=v1y1+v2y2

Where

$$v_1 = \int \frac{-y_2 r}{W(y_1,y_2)}dx$$

and

$$v_2= \int \frac{y_1 r}{W(y_1,y_2)} dx$$


(In this example r=sinx/cosx i.e. the RHS of the DE)


Working it out now


$$v_1= \int \frac{-cosx*\frac{sinx}{cosx}}{-1}dx= \int sinx dx=-cosx+c_1$$



$$v_2= \int \frac{sinx*\frac{sinx}{cosx}}{-1}dx= \int \frac{cos^2x-1}{cosx}dx=\int (cosx-secx)dx=sinx-ln(sex+tanx)+c_2$$


Now putting it back into out solution of y=y1v1+y2v2, we get


y=sinx(-cosx+c1) +cosx(sinx-ln(secx+tanx +c2)

y= c1sinx+c2sinx -sinxcosx-cosxln(secx+tanx)


(Noting that y is always the sum of yc and yp where yc=c1sinx+c2cosx)

 

[alpha754293]

Oh...

I think I get it now. sorta. So you have to start with like some kind of either given or assumed form of the solution for it to work.

I am also guessing that you kinda have to do it this way because there aren't too many options in solving those problems?

(Seems like it's a lot of work to get an answer, but I should really be less surprised considering that the class that I'm supposed to be learning this stuff in is basically PDE with initial and boundary conditions (I don't know if you would consider that PDE/BVP or just BVPs), but in either case, it's that...on steroids. And then some.)

Like I said, I've tired to do some research to solve the problem that was given to us and the prof isn't very helpful in explaining it. Sadly, the one good math prof that I like is a semi-retired adjunct professor and he's pretty much on and off campus at random.

Next dumb question: How do you integrate a Wronskian?