Partial derivatives and transformations of variables: A step-by-step guide.

[Lawrencel2]

Show that under the transformation

u=x, v=$\alpha$x+$\beta$t​

Ay_xx+By_xt+Cy_tt=0 ; B^2-4AC>0
becomes

Ay_uu+(2A$\alpha$+B$\beta$)y_uv+(A$\alpha$²+B$\alpha$$\beta$+C$\beta$²)y_vv=0

(A,B,C are constants)
I have no idea where to start. and i have to present this problem to the front of my class on monday. Can anyone give me a big head start or anything?

 

[jackmell]

I have no idea where to start.

gotta' start taking partials. That's where to start.

and i have to present this problem to the front of my class on monday.

Is it like a bunch of people in there?

Ok, just playing.

So if y=f(x,t) and x=u and v=ax+bt, then:

$$y_x=y_u u_x+y_v v_x$$
$$y_x=y_u+a y_v$$

but the second one is a litle tricky since you taking the partial of partials so:

$$y_{xx}=\frac{\partial}{\partial x} \left(y_u+a y_v\right)=y_{uu} u_x+y_{vu} v_x+a\left(y_{uv}u_x+y_{vv} v_x\right)$$

Ok then, keep doing that for each partial in the first expression, make all those substitutions back into the first expression which wil turn it into an expression of y in terms of u and v, then simplify, cancel, whatever, and it should look like the second expression.