# Math Help - Itegration

1. ## Itegration

Integrate the following equation with resprect to x twice:

f'(x) + x f''(x) = 0

2. Hello, feage7!

A slightly different approach . . .

Integrate with resprect to $x$ twice: . $f'(x) + x f''(x) \:= \:0$

Note that we have: . $\frac{d}{dx}\left[x\!\cdot\!f'(x)\right] \:=\:0$

Integrate: . $x\!\cdot\!f'(x)\:=\:C_1$

Integrate again: . $\int x\!\cdot\!f'(x)\,dx\:=\:\int C_1\,dx$

By parts:
. . $\begin{array}{cc}u = x & dv = f'(x)\,dx \\ du = dx & v = f(x)\end{array}$

And we have: . $x\!\cdot\!f(x) - \int f(x)\,dx\:=\:C_1x + C_2$

3. 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