Having Trouble with this DEQ

  • Thread starter dsr39
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  • #1
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Main Question or Discussion Point

I am trying to solve

[tex]x^2y'' + xy' = 0[/tex]

I know that two solutions that work, by inspection are

[tex]y_1 = c_1[/tex]

and

[tex]y_2 = c_2ln(x)[/tex]

where [tex]c_1[/tex] and [tex]c_2[/tex] are just arbitrary constants.

However I was hoping I would be able to find a more systematic way to solve it. I can get the constant solution by the method of Frobenius, but when I tried to get the natural log solution using the Wronskian, I wasn't able to get it.

I ended up with the second solution being the integral of a Gaussian or something of that sort. I know the Wronskian method is always supposed to work for DEQ's of this type. Where am I going wrong? Can someone show me how to use the Wronskian and the constant solution to get the natural log solution?

(Also why do my LaTeX expressions look funny)

Thanks for the help
 

Answers and Replies

  • #2
HallsofIvy
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Since y itself does not appear in that equation, there is a pretty standard method. Let u= y'. Then the equation becomes x2u'+ xu= 0 which is separable:
x2u'= -xu or (1/u)du= -(1/x)dx. Integrating, ln(u)= -ln(x)+ C or dy/dx= u= c/x (where c= eC) dy= (c/x)dx gives y= cln(x)+ D.

This also happens to be an "Euler-type" or "equi-potential" equation- the power of x in each term is equal to the order of the derivative. The change of variable, u= ln(x), so that x= eu, makes that an equation with constant coefficients: dy/dx= (dy/du)(du/dx)= (1/x)dy/du and d2y/dx2= d/dx(dy/dx)= d/dx((1/x)dy/du)= (1/x)d/du((1/x)dy/du)= (1/x)((1/x)d2y/du2)-(1/x2dy/dx.
The equation becomes d2y/dx2- dy/du+ dy/du= d2/du2= 0. Integrating once y'(u)= C. Integrating again y(u)= Cu+ D. Since u= ln(x), y(x)= Cln(x)+ D.
 
  • #3
tiny-tim
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(Also why do my LaTeX expressions look funny)
Hi dsr39! :smile:

Do you mean in the line "where [tex]c_1[/tex] and [tex]c_2[/tex] are just arbitrary constants."?

Because when it's in the middle of a text line, you need to use "itex" (for "inline latex") instead of "tex" …

where [itex]c_1[/itex] and [itex]c_2[/itex] are just arbitrary constants. :wink:
 
  • #4
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hmm... What if the equation is (x^2)*y"[x]+x*y'[x]-y[x] = 0?

How different would the solution be?
 
  • #5
lurflurf
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x^2*y''+x*y'=0
x*y''+y'=0
(x*y')'=0
x*y'=C1
y'=C1/x
y=C1*log(x)+C2
 
  • #6
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Not sure if u are replying to my question or the original poster's question.
What if there is an additional y[x] term in the equation? i.e.

(x^2)*y"[x]+x*y'[x]-y[x] = 0
 
  • #7
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We could try guessing solutions.

Try y = x. This works.

Indeed, y = cx works, for all constants c.

Does y = cx + d work? not really, since you have that 'd' left over.

So far, we can be sure that the general solution will contain a "cx" term.

Similarly, one can see that a/x works too; try it out.

So I'm guessing the general solution is

y(x) = ax + b/x

Notice that it has two degrees of freedom, as required for a 2nd order ODE.
 
  • #8
lurflurf
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(x^2)*y"+x*y'-y = 0
[(x^2)*y'-x*y]'=0
let y=u/x
[(x^2)*y'-x*y]'=0->x*u''-u'=0
u'=2x*C2
u=C1+(x^2)*C2
y=u/x=C1/x+x*C2
 
Last edited:
  • #9
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Nice.

Does it come out to be what I got?
 
  • #10
lurflurf
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^Now that I fixed my error they match.
 

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