Method of characteristics for linear PDE's (variable coefficients)

Defconist
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I was going through an inroductory book on PDE's and at one point they proceed with little show of work. I have problem with equation -yu_x + xu_y = u.
Characteristics for this equation are x_t = -y, y_t = x, u_t = u.

So far it is clear, but now books states that solution of first characteristic is x(t,s) = f_1(s)sin(t) + f_2(s)cos(t), which is perplexing to me, I would just integrate righthand side treating x or y as constants (we are integrating with respect to t). Any suggestion?
 
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Defconist said:
Characteristics for this equation are
x_t = -y, y_t = x, u_t = u.

So far it is clear, but now books states that solution of first characteristic is x(t,s) = f_1(s)sin(t) + f_2(s)cos(t), which is perplexing to me, I would just integrate righthand side treating x or y as constants (we are integrating with respect to t). Any suggestion?

Hi Defconist! Welcome to PF! :smile:

Nooo … x and y depend on t, so if you vary t, then you must vary x and y … they're not constants!

Hint: xt = -y, yt = x,

means that xtt = -x. :smile:
 
Oh, I get it. It is a system od ODE's because the in y the second equation is the same as y in the first one... It's easy to see why I missed that. It is a possibility I feared from the very beginning. Anyway, thanks for getting me from this predicament. :)
 
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