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APM346-2022S => APM346--Lectures & Home Assignments => Chapter 2 => Topic started by: Xiangmin.Z on January 17, 2022, 04:40:37 PM

Title: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 04:40:37 PM

Hello, I have a question about example 4 from W2 L1:
We are given :$u_{t}+xu_{x}=xt $
after calculation we get:
$x=Ce^t$

$du=xt dt=Cte^t dt$, so $u=C(t-1)e^t+D=x(t-1)+D$,
but how did we get $D=\phi({xe^{-t}} )$? we know it is a constant, but why is D depended on $xe^{-t}$?

Also, why would the initial condition $u|_{t=0} =0 $ implies that $\phi({x}) =x$ ?

Thanks.

Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 05:27:38 PM

I checked the calculation and I think the answer for x is correct, and does anyone know why D is depended on $xe^{-t}$?

Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Victor Ivrii on January 17, 2022, 07:48:38 PM

Now it is correct $x=Ce^{t}$ and then $C=?$

Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 08:19:53 PM

$C=xe^{-t}$, so D is a function of $\phi$, therefore $D=\phi({xe^{-t}} )$ ? Yes, but "$D$  is a function of it"