## Variation of parameters

1. The problem statement, all variables and given/known data

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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 Quote by kasse 1. The problem statement, all variables and given/known data 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 $x^2 y"- 2xy'+ 2=0$. 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.

Recognitions:
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Science Advisor
Staff Emeritus
 Quote by kasse 1. The problem statement, all variables and given/known data 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 $x^2 y"- 2xy'+ 2=0$. 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.

## Variation of parameters

Thanks!
 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.

 Quote by kasse 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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