# Homework Help: Differential equation

1. Dec 22, 2008

### Firepanda

This whole question is to do with a 1D heat equation.

This is where I get stuck

He has derived this far

Q'(t) = -Q(t) * (cb)^2

and

P''(x) = -P(x) * b^2

Now, his differntial equation for Q(t) is

Q(t) = Ae^-(cb)^2 t

and

P(x) = Bcos(bx) + Csin(bx)

How did he get these?

Only way I can think of is for the second one where i would rearrange into a homogeneous equation

P'' + b^2 P = 0

Where P = 0 or -b^2 (from auxillery eqn)

so the general solution would be P(x) = B + Ce^-b^2 x

But this isn't what he found..

2. Dec 22, 2008

### rock.freak667

$$Q'(t) = -Q(t)(cb)^2$$

$$\frac{Q'(t)}{Q(t)}=-(cb)^2$$

$$\int \frac{Q'(t)}{Q(t)}dt= \int -(cb)^2 dt$$

and now d/dt{lnQ(t)}=Q'(t)/Q(t)

For P''(x) = -P(x) * b^2

P''(x)+P(x) * b^2=0

the auxiliary equation would be r2+b2=0
What kind of roots would this give?