How can I determine the correct solutions for a variation of parameters problem?

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Homework Help Overview

The discussion revolves around solving a second-order linear differential equation using the method of variation of parameters. The equation is of the form x^2y'' - 2xy' + 2y = x^3cos(x), and participants are exploring the general solution and specific aspects of the method.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the identification of solutions to the homogeneous equation, questioning whether inspection is the only method. They also explore the validity of certain proposed solutions and the implications of their choices on the particular solution.
  • There are inquiries about the selection of functions for the Wronskian and how this affects the outcome, including confusion regarding the constants of integration in the general solution.

Discussion Status

Participants are actively questioning their understanding of the method, particularly regarding the Wronskian and the choice of solutions. Some guidance has been offered about the nature of the homogeneous equation and the role of constants in the general solution, but no consensus has been reached on the implications of these choices.

Contextual Notes

There is an ongoing exploration of the assumptions regarding the nature of the solutions and the specific requirements of the variation of parameters method. Participants express confusion about the Wronskian and its impact on the particular solution, indicating a need for further clarification.

kasse
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Homework Statement



x^2y''-2xy'+2y=x^3cosx

Find a general solution by using variation of parameters

2. The attempt at a solution

I already solved this one, but I have 4 questions:

1. I found the solutions x and x^2 to the homogeneous equation by inspection. Is this the only way to do it?

2. How about the solutions -x and -x^2? I first tried the solutions x^2 and -x^2, and that gave me 0 as the particular solution...Why can't I choose these two?

3. When calculating the Wronski, how do I know which solution is y1 and y2? I first chose y1 as x^2, and that gave me a different solution...

4. Why are there no constants of integration?
 
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kasse said:

Homework Statement



x^2y''-2xy'+2y=x^3cosx

Find a general solution by using variation of parameters

2. The attempt at a solution

I already solved this one, but I have 4 questions:

1. I found the solutions x and x^2 to the homogeneous equation by inspection. Is this the only way to do it?
The homogeneous equation here is [itex]x^2 y"- 2xy'+ 2=0[/itex]. That's an "equipotential" equation, also called an "Euler type" equation. You can get its general solution by "trying" y= xr as a solution. Also the substitution x= ln(t) changes the equation into a "constant coefficients" equation that is easy to solve.

2. How about the solutions -x and -x^2? I first tried the solutions x^2 and -x^2, and that gave me 0 as the particular solution...Why can't I choose these two?
The general solution to the homogeneous equation is Cx+ Dx2 where C and D can be any contstants. You can choose them to be -1, certainly. I don't know what you mean by "gave me 0 as the particular solution". y identically equal to 0 certainly doesn't satisfy the equation.

3. When calculating the Wronski, how do I know which solution is y1 and y2? I first chose y1 as x^2, and that gave me a different solution...
The Wronskian is a determinant. If you swap two columns, you just multiply the Wronskian by -1. That should do no more than multiply your previous solution by -1.

4. Why are there no constants of integration?
Because youdidn't put them in! Your general solution should be Cx+ Dx2+ a particular solution to the entire equation.
 
kasse said:

Homework Statement



x^2y''-2xy'+2y=x^3cosx

Find a general solution by using variation of parameters

2. The attempt at a solution

I already solved this one, but I have 4 questions:

1. I found the solutions x and x^2 to the homogeneous equation by inspection. Is this the only way to do it?
The homogeneous equation here is [itex]x^2 y"- 2xy'+ 2=0[/itex]. That's an "equipotential" equation, also called an "Euler type" equation. You can get its general solution by "trying" y= xr as a solution. Also the substitution x= ln(t) changes the equation into a "constant coefficients" equation that is easy to solve.

2. How about the solutions -x and -x^2? I first tried the solutions x^2 and -x^2, and that gave me 0 as the particular solution...Why can't I choose these two?
The general solution to the homogeneous equation is Cx+ Dx2 where C and D can be any contstants. You can choose them to be -1, certainly. I don't know what you mean by "gave me 0 as the particular solution". y identically equal to 0 certainly doesn't satisfy the equation.

3. When calculating the Wronski, how do I know which solution is y1 and y2? I first chose y1 as x^2, and that gave me a different solution...
The Wronskian is a determinant. If you swap two columns, you just multiply the Wronskian by -1. That should do no more than multiply your previous solution by -1.

4. Why are there no constants of integration?
Because youdidn't put them in! Your general solution should be Cx+ Dx2+ a particular solution to the entire equation.
 
Thanks!
 
Last edited:
The Wronski keeps confusing me, however.

W(x^2, x) = -x^2

W(x, x^2) = x^2

The particular solution depends on which one of these I choose.
 
kasse said:
The Wronski keeps confusing me, however.

W(x^2, x) = -x^2

W(x, x^2) = x^2

The particular solution depends on which one of these I choose.

Or not?
 

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