Cauchy Euler, non-homogeneous, weird condition

In summary: However, since it was my question, it seemed more fitting to post it here. I apologize for the double post. In summary, the conversation involved solving a non-homogeneous Cauchy-Euler equation using variation of parameters, with the given conditions of y(1)=0 and y(0) being bounded. The incorrect method of solving was shown, and then the correct method was explained. The conversation also touched on the use of different methods for variation of parameters.
  • #1
ericm1234
73
2
xy''+y'=-x
y(1)=0, y(0) bounded (so the natural log, 1/x etc. terms drop out)
homogeneous, cauchy euler: y=a+bx
variation of parameters, and using the conditions gives y=1-x, I think (i tried this previously and I think this is what I got, I didnt write it down). Very different from what I have on my solution handout.
First off, does it matter when to apply the bounded condition? (as in, do we drop the blnx term first? the problem here is variation of parameters doesn't work then),
Is there another way to solve a non-homogeneous cauchy euler besides using variation of parameters?
 
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  • #2
hi ericm1234! :smile:
ericm1234 said:
xy''+y'=-x

in this particular case, isn't the LHS obviously a perfect derivative? :wink:
 
  • #3
Man..thanks Tim I see that now. Very easy once that is spotted.
I tried over and over to solve this as a cauchy euler, then using variation of parameters; would that approach be invalid? I couldn't get the answer.
 
  • #4
I get the homogeneous solution y=a+blnx
variation of parameters gives u1= x^3/3-x^3/9 and u2=-x^3/3, y=a+blnx+u1+u2lnx
Different from what I get (the correct answer) the other way.
Help.
 
  • #5
I should add, again: y(1)=0, and y(0) bounded, are the conditions.
Correct answer should be 1/4(1-x^2), so again why isn't my cauchy euler way working
 
  • #6
If you want someone to point out what you did wrong, then you will have to show what you did! I did this problem using "variation of parameters" and I certainly got nothing involving "[itex]x^3[/itex]".
 
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  • #7
I see what I did wrong; my method of variation of parameters requires first dividing by the lead coefficient before going to solve for the particular solution. All good now.
Thanks.
 
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  • #8
Cauchy euler non-homogeneous problem

Homework Statement


Must solve by cauchy euler, with variation of parameters:
Original problem: xy''+y'=-x, y(1)=0, y(0) bounded (which I guess means all 1/x, lnx terms etc drop out)

The Attempt at a Solution



homogeneous: xy''+y'=0 make substitution y=x^r, so homogeneous soln is y=a+blnx (call it yc)
variation of parameters:
u1'=a determinant with entries (from left to right, top to bottom) 0, lnx, -x, 1/x then divide that by the wronskian= 1/x
u2'= a determinant with entries (from left to right, top to bottom) 1, 0, 0 -x, then divide by wronskian= 1/x
u1= x^3/3*lnx-x^3/9 u2=-x^3/3
yp=u1*1+u2*lnx
y=yc+yp

But this is wrong; what am I doing wrong?
 
  • #9


ericm1234 said:

Homework Statement


Must solve by cauchy euler, with variation of parameters:
Original problem: xy''+y'=-x, y(1)=0, y(0) bounded (which I guess means all 1/x, lnx terms etc drop out)

The Attempt at a Solution



homogeneous: xy''+y'=0 make substitution y=x^r, so homogeneous soln is y=a+blnx (call it yc)
Okay, the characteristic equation is [itex]r^2= 0[/itex] which has r= 0 as a double root. That gives the general solution to the associated homogeneous equation, y= a+ b ln(x).

variation of parameters:
u1'=a determinant with entries (from left to right, top to bottom) 0, lnx, -x, 1/x then divide that by the wronskian= 1/x
u2'= a determinant with entries (from left to right, top to bottom) 1, 0, 0 -x, then divide by wronskian= 1/x
u1= x^3/3*lnx-x^3/9 u2=-x^3/3
yp=u1*1+u2*lnx
y=yc+yp

But this is wrong; what am I doing wrong?
Using "variation of parameters", we look for a solution of the form y(x)= u1(x)ln(x)+u2(x). Then y'= u1'ln(x)+ u1/x+ u2'. Because there are, in fact, an infinite number of such solutions, we narrow the search and simplify by requiring that u1'ln(x)+ u2'= 0. Then y'= u1/x so that [itex]y''= u'/x- u/x^2[/itex].

With that, [itex]xy''+ y'= x(u1'/x- u1/x^2)+ u1/x= u1'= -x[/itex]. Integrating that [itex]u1= -x^2/2[/itex]. [itex]u1'ln(x)+ u2'= -xln(x)+ u2'= 0[/itex] gives us [itex]u2'= xln(x)[/itex]. That is NOT going to give your u1 and u2.
 
  • #10


Thanks.
I learned a different method (or different way of viewing it anyway) for variation of parameters, and actually it required diving out the lead coefficient before beginning to solve for the particular solution. Everything's good now.
 
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  • #11
ericm1234, you also posted this under "homework" so I am going to merge the two threads there.

Double posting is a violation of the forum rules and can result in penalties and a possible ban from the forum.
 
  • #12
I reposted it there as it seems more of a "coursework" relevant post.
 

1. What is the Cauchy Euler equation?

The Cauchy Euler equation is a second-order linear differential equation that takes the form an(x)y(n)(x) + an-1(x)y(n-1)(x) + ... + a0(x)y = 0, where the coefficients an(x), an-1(x), ..., a0(x) are functions of x and y(n)(x) represents the nth derivative of y with respect to x.

2. What does it mean for the Cauchy Euler equation to be non-homogeneous?

A non-homogeneous Cauchy Euler equation is one in which the right side of the equation is not equal to zero. In other words, there is a non-zero function on the right side that affects the behavior of the solution.

3. Can the Cauchy Euler equation have weird conditions?

Yes, the Cauchy Euler equation can have weird conditions, such as boundary conditions or initial conditions that are unusual or unexpected. These conditions can complicate the solution and require a different approach to solving the equation.

4. How do you solve a Cauchy Euler equation with weird conditions?

The approach to solving a Cauchy Euler equation with weird conditions depends on the specific conditions and the complexity of the equation. In some cases, it may be necessary to use special techniques, such as Green's functions or Laplace transforms, to find a solution.

5. Why is the Cauchy Euler equation important in mathematics?

The Cauchy Euler equation is important because it is a fundamental tool in many areas of mathematics and physics. It is used to model various phenomena, such as vibrations, electrical circuits, and heat transfer. The solutions to this equation also have many applications in engineering, science, and technology.

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