Thread: 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!

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

First and Second Order Differential Equations

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.

By the null factor law

Now the solution to

Now

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.

This makes things more difficult.

Have a look at this link

solve y''+(2/x)y' =0 - Wolfram|Alpha

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, so that where . Going back to V we have and that's an easy integral.