Second Order ODE with Weird Coefficients: Solving the Equation x2-2x+1

[Design]

Homework Statement

y''(x-1)-xy'+y=x²-2x+1

The Attempt at a Solution

I don't even know how to start this, Don't know what substitution to do.

 

[hunt_mat]

It's a linear solution. Have you tried the Frobenius method?

 

[Dickfore]

Is x - 1 a factor that multiplies y'' or is it the argument of that function?

 

[Design]

It's a linear solution. Have you tried the Frobenius method?

Never heard of anything like this. My prof like putting problems we haven't learned in the problem set = ='. I guess its time to look it up.

Is x - 1 a factor that multiplies y'' or is it the argument of that function?

Don't know what you mean but y'' mutipled by (x-1)

 

[HallsofIvy]

What you wrote originally, y"(x- 1) looked like "f(x-1)" where x- 1 is the argument of the function.

Frobenius' method consists of looking for series solution of the form \sum_{n=0}^\infty a_nx^{n+c} where c does not have to be a positive integer.

 

[Design]

Is there any other way of doing it? It just seems weird because I haven't covered anything about series solutions and all of a sudden this comes up.

 

[Saladsamurai]

Is there any other way of doing it? It just seems weird because I haven't covered anything about series solutions and all of a sudden this comes up.

Doubtful. We have covered everything up until Frobenius method and I cannot think of any other way to solve it. As a matter of fact, this is a non-homogenous Frobenius EQ which we have not covered ... but that just makes more work, not necessarily "harder" work. Give it a try! You'll be ahead of the curve. I'll get you started. As Halls said, seek a solution in the form:

y = \sum_0^\infty a_nx^{n+s}\qquad(1)

\Rightarrow y' = \sum_0^\infty (n+s)a_nx^{n+s-1}\qquad(2)

\Rightarrow y' = \sum_0^\infty (n+s)(n+s-1)a_nx^{n+s-2}\qquad(3)

Plug EQs 1 - 3 back into the original DE and see what happens. But be prepared, Frobenius solutions take awhile and usually for me at least 2 pages of work.

 

[Design]

Doubtful. We have covered everything up until Frobenius method and I cannot think of any other way to solve it. As a matter of fact, this is a non-homogenous Frobenius EQ which we have not covered ... but that just makes more work, not necessarily "harder" work. Give it a try! You'll be ahead of the curve. I'll get you started. As Halls said, seek a solution in the form:

y = \sum_0^\infty a_nx^{n+s}\qquad(1)

\Rightarrow y' = \sum_0^\infty (n+s)a_nx^{n+s-1}\qquad(2)

\Rightarrow y' = \sum_0^\infty (n+s)(n+s-1)a_nx^{n+s-2}\qquad(3)

Plug EQs 1 - 3 back into the original DE and see what happens. But be prepared, Frobenius solutions take awhile and usually for me at least 2 pages of work.

Thanks, will definitely give it a shot.

 

[PAllen]

There is possibly another way to solve this. It is linear, second order, non-constant coefficients. It should be covered somewhere in any reasonable ODE book. You find two linearly independent solutions of the homogeneous equation:

y''(x-1)-xy'+y=0

Then you use these solutions to construct a determinant called a Wronskian. There are then then mechanical procedures to construct solutions of the original equation. I can't give all the details (I'm not going to type in 3 pages), but there should be several pages or a small chapter covering this method in any ODE book.

 

[Saladsamurai]

There is possibly another way to solve this. It is linear, second order, non-constant coefficients. It should be covered somewhere in any reasonable ODE book. You find two linearly independent solutions of the homogeneous equation:

y''(x-1)-xy'+y=0

Then you use these solutions to construct a determinant called a Wronskian. There are then then mechanical procedures to construct solutions of the original equation. I can't give all the details (I'm not going to type in 3 pages), but there should be several pages or a small chapter covering this method in any ODE book.

This is called variation of parameters. I thought you could apply it to nonconstant coefficient problems, but I am not so sure. We never actually have. Now the problem becomes, how do you solve the homogenous equation with nonconstant coefficients to get your homogenous solution so that you can apply variation?

 

[jackmell]

Homework Statement

y''(x-1)-xy'+y=x²-2x+1

The Attempt at a Solution

I don't even know how to start this, Don't know what substitution to do.

Design, I can think of only one way to solve this problem in a classy manner: you have got to go out of your way to go to the library and find a decent ordinary DE textbook and go through the example of "reduction of order" but first it would be nice if you learn when you have problems like this to just quickly see if "simple" solutions work for the homogeneous case like a constant, x, x^2, wait, isn't x a solution to the homogeneous case? The second derivative of x is zero and the rest is just -x+x=0, bingo-bango. Then the next step in reduction of order is to let:

y=xv

and just turn the crank.

And you should have written it as:

(x-1)y''-xy'+y=x^2-2x+1

 

[HallsofIvy]

There is possibly another way to solve this. It is linear, second order, non-constant coefficients. It should be covered somewhere in any reasonable ODE book. You find two linearly independent solutions of the homogeneous equation:

y''(x-1)-xy'+y=0

Then you use these solutions to construct a determinant called a Wronskian. There are then then mechanical procedures to construct solutions of the original equation. I can't give all the details (I'm not going to type in 3 pages), but there should be several pages or a small chapter covering this method in any ODE book.

Yes, of course, but the problem is finding the general solution to the associated homogenous equation and you seem to be taking that for granted!

 

[PAllen]

Yes, of course, but the problem is finding the general solution to the associated homogenous equation and you seem to be taking that for granted!

Not taking anything for granted, the OP asked how to get started. A big first step is to break it into simpler problems. Also, I was disagreeing with other posters that there was no reasonable method except series. I would first try the homogenous equation, before trying series approaches. In this case, as jackmell has already observed, there is at least one trivial solution the homogenous equation.

Also, as to the claim that this method might not be valid, that is silly. Back when I did this stuff (40 years ago), I routinely solved using this method for non-constant coefficients.

The approach I am referring to is just a special case of what jackmell is referring to: specializing the more general reduction of order techniques to a recipe for second order linear ODE with non-constant cefficients.

 

[Dickfore]

Homework Statement

y''(x-1)-xy'+y=x²-2x+1

The Attempt at a Solution

I don't even know how to start this, Don't know what substitution to do.

I will assume that the factor (x - 1) is a multiplier in front of y''. You may rewrite the equation as:

<br /> (x - 1) \, y&#039;&#039; - (x - 1 + 1) \, y&#039; + y = (x - 1)^{2}<br />

<br /> (x - 1) (y&#039;&#039; - y&#039;) - (y&#039; - y) = (x - 1)^{2}<br />

If you define:

<br /> z \equiv y&#039; - y<br />

you get a first order inhomogeneous linear differential equation:

<br /> z&#039; - \frac{1}{x - 1} \, z = x - 1<br />

You can find the general solution of this equation which will contain one arbitrary integrating constant. After that, you may substitute in:

<br /> y&#039; - y = z(x)<br />

which is another equation of the same type and you may find its general solution as well. This will give you one more integrating constant.

 

[jackmell]

I will assume that the factor (x - 1) is a multiplier in front of y&#039;&#039;. You may rewrite the equation as:

<br /> (x - 1) \, y&#039;&#039; - (x - 1 + 1) \, y&#039; + y = (x - 1)^{2}<br />

<br /> (x - 1) (y&#039;&#039; - y&#039;) - (y&#039; - y) = (x - 1)^{2}<br />

If you define:

<br /> z \equiv y&#039; - y<br />

you get a first order inhomogeneous linear differential equation:

<br /> z&#039; - \frac{1}{x - 1} \, z = x - 1<br />

You can find the general solution of this equation which will contain one arbitrary integrating constant. After that, you may substitute in:

<br /> y&#039; - y = z(x)<br />

which is another equation of the same type and you may find its general solution as well. This will give you one more integrating constant.

Ok, sorry. Two. :)

 

[Design]

I will assume that the factor (x - 1) is a multiplier in front of y&#039;&#039;. You may rewrite the equation as:

<br /> (x - 1) \, y&#039;&#039; - (x - 1 + 1) \, y&#039; + y = (x - 1)^{2}<br />

<br /> (x - 1) (y&#039;&#039; - y&#039;) - (y&#039; - y) = (x - 1)^{2}<br />

If you define:

<br /> z \equiv y&#039; - y<br />

you get a first order inhomogeneous linear differential equation:

<br /> z&#039; - \frac{1}{x - 1} \, z = x - 1<br />

You can find the general solution of this equation which will contain one arbitrary integrating constant. After that, you may substitute in:

<br /> y&#039; - y = z(x)<br />

which is another equation of the same type and you may find its general solution as well. This will give you one more integrating constant.

After trying this, I got a integral e^-x/(x-1) which is unsolvable.

 

[Dickfore]

Are you sure the factor (x - 1) is in the denominator? Actually, I think your solution for z is completely wrong.