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Chapter 2 / Re: Week 2 Lec 1 (Chapter 2) question
« 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"
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Chapter 2 / Re: Week 2 Lec 1 (Chapter 2) question
« 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 $xe^{-t}$?
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Chapter 2 / Week 2 Lec 1 (Chapter 2) question
« 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.
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.
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