Method of Characteristics 3: The general case

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For anyone confused, the last result on the page at 11:11 should read -1/5 outside the brackets rather than 1/5.

oliveraherne
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you're a lifesaver man, good vids

bryano
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explaint more about why you get x(0) =s, and y(0)=0?. thanks

daohung
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At 7:34, I would recommend using the root method and the method of underdetermined coefficients instead. This lets you avoid doing all the annoying integration by parts.

ALternaprof
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Thanks a lot for providing simple explanations for these cases.

sushants
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you dont really say how this relates to the others... 1 you paramatize one you just know the slope... shouldn't this be called a different method all together?

tag_of_frank
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How about the 2nd-order ones? I know that at least the wave equation can be solved with characteristics.

bonbonpony
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find so difficult to understand how you get c(x) and the integration of each colorful terms

rythemofgermay
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Thx alot my friend. What did you study if im allowed to ask? :=)

Lyradopon
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i cannot solve the question. can u help me please?


Given the linear equation: Ux-Uy=0 with the initial conditions: x(0, s)=0, y(0, s)=s, u(0, s)=g(s) where g(s) is an arbitrary differentiable function.
a) Write the Characterist equations: a(x, y, u)Ux+b(x, y, u)Uy=c(x, y, u)
dx = a(x, y, u) (1) dy = b(x, y, u) (2) du = c(x, y, u) (3)
dt dt dt

b) Integrate equations (1-2), use the initial conditions and determine x and y interms of the parameters t and s, then inverting these, write t and s interms of x and y.
c) Integrate equations (3), use the initial conditions and determine u interms of t and s and then write u interms of x and y:u(x, y)

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