# Second Order ODE -Euler

• Jul 22nd 2012, 04:30 PM
Tribal5
Second Order ODE -Euler
I'm trying to work out the following Cauchy-Euler problem, was hoping someone could help
the equation is:

1/2*J*s^2 v'' + (r -D)S v' - rV =0, for 0<S<A

V(s) = S - E, for s>a

when V(0) = 0 , V(A) = A - E, dV/dS = 1

I've to show v(s) = (A - E)(S/A)^m

where m = 1/J * [-(r-D-J/2) + Sqrt( (r-D-J/2)^2 + 2rJ))

note J = Sigma^2 (Sorry don't know how to use LaTeX)
I can work out m using V = Sm and i guess the root is v = C1Sm + C2Sm

But i just can't tie it all together, any ideas?
• Jul 22nd 2012, 06:24 PM
Prove It
Re: Second Order ODE -Euler
Quote:

Originally Posted by Tribal5
I'm trying to work out the following Cauchy-Euler problem, was hoping someone could help
the equation is:

1/2*J*s^2 v'' + (r -D)S v' - rV =0, for 0<S<A

V(s) = S - E, for s>a

when V(0) = 0 , V(A) = A - E, dV/dS = 1

I've to show v(s) = (A - E)(S/A)^m

where m = 1/J * [-(r-D-J/2) + Sqrt( (r-D-J/2)^2 + 2rJ))

note J = Sigma^2 (Sorry don't know how to use LaTeX)
I can work out m using V = Sm and i guess the root is v = C1Sm + C2Sm

But i just can't tie it all together, any ideas?

Are you using S and s to represent the same thing? If so, please be consistent...
• Jul 22nd 2012, 11:38 PM
Tribal5
Re: Second Order ODE -Euler
Yes sorry s and S are the same I typed a bit quick.
• Jul 23rd 2012, 02:17 AM
Prove It
Re: Second Order ODE -Euler
Quote:

Originally Posted by Tribal5
I'm trying to work out the following Cauchy-Euler problem, was hoping someone could help
the equation is:

1/2*J*s^2 v'' + (r -D)S v' - rV =0, for 0<S<A

V(s) = S - E, for s>a

when V(0) = 0 , V(A) = A - E, dV/dS = 1

I've to show v(s) = (A - E)(S/A)^m

where m = 1/J * [-(r-D-J/2) + Sqrt( (r-D-J/2)^2 + 2rJ))

note J = Sigma^2 (Sorry don't know how to use LaTeX)
I can work out m using V = Sm and i guess the root is v = C1Sm + C2Sm

But i just can't tie it all together, any ideas?

Why would you guess \displaystyle \displaystyle \begin{align*} C_1s^m + C_2s^m \end{align*}? That simplifies to \displaystyle \displaystyle \begin{align*} \left(C_1 + C_2\right)s^m = C\,s^m \end{align*}. So you only need to guess a single exponential function.

Let \displaystyle \displaystyle \begin{align*} v = C\,s^m \implies v' = Cm\,s^{m - 1} \implies v'' = Cm(m-1)s^{m-2} \end{align*}.

Substitute these into your DE and see if you can solve...
• Jul 23rd 2012, 03:03 PM
Tribal5
Re: Second Order ODE -Euler
yes solving gives the quadratic m^2 + (2(r-D)/J - 1) m -2r/J = 0
if i say b^2 - 4ac > 0
the roots will be
then m(+ve) = 1/J * [-(r-D-J/2) + Sqrt( (r-D-J/2)^2 + 2rJ))

solution would be A S^m(+ve) + B S^m(-ve)

however i've to move from this to V(S) = (A - E)(S/A)^m(+ve) - sorry should have put +ve and -ve in