Solving 2nd Order ODE: y''+(1/x)y'=0

[sara_87]

Homework Statement

find the general solution to the ODE:
y''+(1/x)y'=0

Homework Equations

The Attempt at a Solution

I put this in the following form: y''=-(1/x)y'
integrated both sides: y'=-ln(x)y +C
I think i made a stupid mistake but i can't figure out what it is.
Any help would be appreciated. Thank you.

 

[phreak]

You can't integrate like that. You can only integrate when you have an ODE of the form y' = f(x), where anything y isn't on the right side.

There is a general solution to ODEs of the form y'' + a(x)y' = 0, and it's probably in your textbook. Hint: multiply both sides by [tex]e^{\int^x_{x_0} a(t)dt}[/tex], and contract by the product rule for differentiation.

 

[sara_87]

why do i have to multiply both sides by that integral?

 

[phreak]

Because it gives you something you can integrate. You're trying to find y, right? Well, if you multiply both sides by the given function, and note that the derivative of [tex]e^{\int^x_{x_0} a(t)dt}[/tex] is [tex]a(x)e^{\int^x_{x_0} a(t)dt}[/tex], you can use the product rule to write the equation in the form (X y')' = 0, where X is something you need to find yourself (I've given you all the clues you need). You can then conclude that X y' = c, for some constant c, by integrating both sides. You now have an equation that is easily solvable if you know the very basics of differential equations.

 

[sara_87]

ok, so i did this and my working out is:
so the 'a' is 1/x, and integrating this gives lnx so e^ln(x)=x so multiplying the ODE vy x gives:
xY''+Y'=0
so (xY')'=0
integrating both sides gives:
xY'=C (where C is a constant)
integrating again gives:
Y=Cln(x)+D (where D is a constant)
Is this correct?