Formula that gives all possible solutions

Hi there,

I'm working on a problem that asks me to find a formula that'll give all possible solutions to the equation V'' + (2/3)V'(r) = 0. I'm not certain where to begin. I believe that solving for V'(r) would be a nice first step, but I don't know how to go about this either. ANy guidance would be greatly appreciated. Thank you!

Re: Formula that gives all possible solutions

Look at this link under "Second Order Differential equations", the first example will show you how to proceed.

First and Second Order Differential Equations

Re: Formula that gives all possible solutions

Thank you for the link pickslides.

From what I've gotten, the solution should be something akin to y = c(sub 1)e^(r(sub1)x) + c(sub 2)e^(r(sub 2)x). The solution to r^2 - (2/r)r = 0 is plus or minus square root of 2. The solution, then, appears to be y = c(sub 1)e^(sqrt(2)x) + c(sub 2)e^(-sqrt(2)x). Is this correct? Thank you once again for your assistance.

Re: Formula that gives all possible solutions

$\displaystyle \displaystyle \lambda^2+\frac{2}{3}\lambda = 0 $

$\displaystyle \displaystyle \lambda\left( \lambda+\frac{2}{3}\right) = 0 $

By the null factor law $\displaystyle \lambda = 0, \frac{-2}{3}$

Now the solution to $\displaystyle V(r)$

Now $\displaystyle \displaystyle V(r) = C_1e^{\frac{-2r}{3}}+C_2e^{0}$

Re: Formula that gives all possible solutions

pickslides, I'm so sorry. I typed the coefficient (2/3) in the original problem when the problem states (2/r). I typed my attempt at a solution with regard to that. I apologize sincerely for my mistake. You've been very kind with your explanation, and I thank you.

Re: Formula that gives all possible solutions

Re: Formula that gives all possible solutions

Quote:

Originally Posted by

**reliance** Hi there,

I'm working on a problem that asks me to find a formula that'll give all possible solutions to the equation V'' + (2/3)V'(r) = 0. I'm not certain where to begin. I believe that solving for V'(r) would be a nice first step, but I don't know how to go about this either. ANy guidance would be greatly appreciated. Thank you!

The equation is V''+ (2/r)V'= 0. Since V itself does not appear in the equation, let U= V' so the equation becomes U'+ (2/r)U= 0. That is a separable first order equation. We can write dU/dr= -(2/r)U and then dU/U= -(2 dr/r). Integrating both sides, $\displaystyle ln(U)= -2ln(r)+ C= ln(r^{-2}+ C$ so that $\displaystyle U= C' r^{-2}$ where $\displaystyle C'= e^C$. Going back to V we have $\displaystyle \frac{dV}{dr}= C'r^{-2}$ and that's an easy integral.