Integrate the following equation with resprect to x twice:
f'(x) + x f''(x) = 0
ok ive tried using that and it didnt help much in terms of what i needed to do so ill try n show the full question, it is a dimensional analysis question but its the integration part im having difficulty on;
you have the following variables: t, R, p, M, e, y, u.
using bukcinghams pi theorem u get the equation
u/ ((t/p)^1/2)=f(Rp/M x (t/p)^1/2 , y/R , e/R)
then by adapting a new constraint another equation is formed basically the same as above only e/R is removed and y/R is now known as n and Rp/M x (t/p)^1/2 is known as E :
u/ ((t/p)^1/2)=f1(En)
then another equation is derived:
f3(E)=f1(En)+f2(n)
differentiate once with respect to E, and then differentiate that answer with respect to n to get the equation
f1'(En) + En f1"(En)=0
let x = En
and integrate twice with respect to x to obtain the equation:
u/((t/p)^1/2)= A ln{y/m x p (t/p)^1/2 } + B
dnt know if ive gave you enough information there