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

In summary, the conversation discusses finding the general solution to the equation x^2y''-2xy'+2y=x^3cosx using variation of parameters. It covers methods for solving the homogeneous equation and choosing appropriate solutions, as well as calculating the Wronskian and understanding its relationship to the particular solution. The conversation also touches on the presence of constants of integration in the general solution.
  • #1
kasse
384
1

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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  • #2
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.
 
  • #3
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.
 
  • #4
Thanks!
 
Last edited:
  • #5
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.
 
  • #6
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?
 

1. What is the concept of "variation of parameters" in mathematics?

The variation of parameters is a method used in solving linear differential equations, particularly in cases where the coefficients of the equation are not constant. It involves finding a particular solution by varying the parameters of a general solution to the homogeneous equation.

2. How is the variation of parameters method different from other methods of solving differential equations?

The variation of parameters method is different from other methods, such as the method of undetermined coefficients or the method of integrating factors, because it allows for the solution of non-homogeneous equations with variable coefficients. It is also more flexible in that it can be used to solve a wider range of equations.

3. What are the steps involved in using the variation of parameters method?

The variation of parameters method involves the following steps: 1) finding a general solution to the corresponding homogeneous equation, 2) finding a particular solution by varying the parameters of the general solution, and 3) combining the general and particular solutions to obtain the complete solution to the non-homogeneous equation.

4. In what situations is the variation of parameters method most useful?

The variation of parameters method is most useful when dealing with differential equations with variable coefficients, as well as equations that cannot be solved using other methods, such as undetermined coefficients or integrating factors. It is also useful in situations where the particular solution is known or can be easily guessed.

5. Are there any limitations to the variation of parameters method?

While the variation of parameters method is a powerful tool for solving differential equations, it does have some limitations. It can only be applied to linear differential equations, and it can be more time-consuming and complex compared to other methods. In some cases, it may also yield a solution that is not in a simplified form.

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