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

[Defconist]

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?

 

[tiny-tim]

Welcome to PF!

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!

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

Hint: x_t = -y, y_t = x,

means that x_tt = -x.

 

[Defconist]

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. :)