2nd Order Linear homog ODE of variable coefficients

[linda300]

Hey,

Every where I look I can only find books and pdf talking about the uniqueness and linear independence of the solutions but I haven't been able to find a procedure of finding the solutions to one of these ode's if you haven't been already given a particular solution.

I've been trying to do this tute question,

http://img848.imageshack.us/img848/8382/sdfsq.jpg

but in my lecture note we only went through the proofs of how the solutions are linearly independent and so on,

Would anyone be able to point we in the right direction to where I can find some examples of ODE's like this, solved without already knowing a particular solution?

I have spent some time trying to guess one of the solutions so I could find the other but it isn't going so well,

Mathematica spits out ((1 + x^2) C[1])/(2 x) + (i(-1 + x^2) C[2])/(2 x)
as the general solution but without knowing how it got there it doesn't really help,

Thanks

 

[HallsofIvy]

If you multiply through by x^2 you get the "Cauchy type" or "equi-potential" equation x^2y''+ xy'- y= 0.

One method of solving that is to try y= x^r. Put that into the equation and see if there exist values of r so that x^r satisfies the equation.

Another method is to change the independent variable: let t= ln(x). That changes the equation to one with constant coefficients.

 

[linda300]

Oh cool thanks!

I got r = ±1

so x and x^-1 are the solutions and they're definitely linearly independent!

There is another aspect of the question, it states initial conditions that must be satisfied,

x satisfies one set however x^-1 doesn't satisfy the other, the other initial condition correspond to ln(x),

Is it possible to find the solution ln(x) using the x^-1 found using x^r as the solution?

 

[HallsofIvy]

I'm not sure what you mean by "x satisifies one set however x^-1 doesn't satisfy the other". They aren't supposed to satisfy one or the other condition separately.

The fact that x and x^-1 satisfy the equation and are independent tells you that any solution to that equation can be written in the form y= C_1 x+ C_2x^{-1}. Replace x and y by the values given in your two conditions to get two equations to solve for C₁ and C₂.

 

[linda300]

Isn't that y = C x + B x^-1 the general solution?

Made up of two solutions y₁ = Cx and y²=Bx^-1

So in the next question it says find the solutions such that

y₁(1) = 1 and y'₁(1) = 0 which works with x

And y₂(1) = 0 and y'₂(1) = 1 which works if y₂=ln(x)

are satisfied,

Would you obtain the ln(x) solution by using the known solution x?

Another thing that I noticed is that ln(x) isn't actually a solution, but what other functions would satisfy the y₂ solution?