# Math Help - Differentiation Crisis

1. ## Differentiation Crisis

Hi I'm stuck with a problem, I'm fine with differentials but the way this is worded confuses me!

Valuation of a storage facility gives equation:

V'' - rV = cP where c and r are constants

V: R->R is a function of the variable P and it wants me to find the general solution V (P).

I'm going round in circles here, if V is a function of P then am I right in saying V'' = d^2V/dP^2 ??
but then where do I go, do they want V = P something?

Hi I'm stuck with a problem, I'm fine with differentials but the way this is worded confuses me!

Valuation of a storage facility gives equation:

V'' - rV = cP where c and r are constants

V: R->R is a function of the variable P and it wants me to find the general solution V (P).

I'm going round in circles here, if V is a function of P then am I right in saying V'' = d^2V/dP^2 ??
but then where do I go, do they want V = P something?

This is a standard second order linear ordinary differential equation.

We first look for the general solution of:

$
V'' - rV = 0
$

For this take a trial solution $V(P)=e^{\lambda P}$, then as:

$
V'' - rV =0
$

we have:

$
\lambda^2 V(P)-rV(P)=0
$

so if $V(P) \not \equiv 0$

$
\lambda^2-r=0
$

and so $\lambda= \pm \sqrt{r}$

So the general solution to $V'' - rV =0$ is:

$
V(P)=A e^{\sqrt{r}P}+B e^{-\sqrt{r}P}\ \ \ \ \dots(1)
$

where $A$ and $B$ are arbitary constants.

Now to get the general solution to $V'' - rV = cP$ we take any solution of this equation and add on the right hand side of $(1)$ above. An obvious solution is V(P)=cP, so the general solution is:

$
V(P)=cP + A e^{\sqrt{r}P}+B e^{-\sqrt{r}P}\ \ \ \ \dots(2)
$

RonL