Use separation of variables to find particular solutions

[Dragonfall]

"Use separation of variables to find particular solutions of

$$u_t-u_{xx}-2u_x=0, 1<x<2, 0<t, u(1,t)=u(2,t)=0$$

hint: change coordinates"

I can't find the solution. The equation seems already separated, so all I need to do is to find a change of variables, I think. But I can't find one that works.

 

[Fermat]

You're looking at it the wrong way.
It's not a change of variables that's wanted, but separation of variables.

Remember this ?

$$\frac{dy}{dx} = \frac{y}{x}$$

separate the variables

$$\frac{dy}{y} = \frac{dx}{x}$$

now integrate
$$\int\frac{dy}{y} = \int\frac{dx}{x}$$
$$ln(y) = ln(x) + c$$

You are expected to do something similar.

First of all, you assume a solution of the form u(x,t) = f(x)g(t)
Now substitute that into your original pde and apply separation of variables.
You should end up with functions of x on one side of the equals sign and functions of t on the other side.
Since the lhs is a function of x and the rhs is a function of t, then the only way they (the two functions) can be equal to each other is to set them both equal to the same constant value. i.e. lhs = C and rhs = C
You now have two independent pdes, (in fact, now they are ode's) one in x and one in t, that you can solve separately.

 

[HallsofIvy]

"Use separation of variables to find particular solutions of

$$u_t-u_{xx}-2u_x=0, 1<x<2, 0<t, u(1,t)=u(2,t)=0$$

hint: change coordinates"

I can't find the solution. The equation seems already separated, so all I need to do is to find a change of variables, I think. But I can't find one that works.

Then you need to review what "separation of variables" means for partial differential equations.

Write u(x,t)= X(x)T(t). That is, X is a function of x only, T a function of t only. The the differential equation becomes
XT'- TX"- 2TX'= 0. Divide both sides by XT to get
$$\frac{T'}{T}- \frac{X"}{X}- \frac{2X'}{X}= 0$$
or
$$\frac{T'}{T}= \frac{X"- 2X'}{X}$$
Since the left side depends on t only and the right side depends on x only the only way they can be equal (for all x and t) is if each side is a constant:
$$\frac{T'}{T}= \alpha$$
so that $T'= \alpha T$
and
$$\frac{X"- 2X'}= \alpha$$
so that $X"- 2X'= \alpha X$
where $\alpha$ is some unknown constant. The general solution will typically be a sum of such things.