Respuesta :
Answer:
We can find the solution using chain rule,
For instance, if u(x,y) = f(ax+by), then
[tex]u_{x}(x,y) = a f^{'} (ax+by)[/tex] which represents derivative of x with respect to x
and then derivative of [tex]u_{x}(x,y)[/tex] with respect to y is
[tex]u_{xy}(x,y) = (ab)^2 (f^{''} (ax+by))^2[/tex],
Now, the derivative of [tex]u_{x}(x,y)[/tex] with respect to x, which is the second derivative, which is
[tex]u_{xx}(x,y) = a^2 f^{''} (ax+by)[/tex]
and the derivative [tex]u_{y}(x,y)[/tex] and [tex]u_{yy}(x,y)[/tex] are
[tex]u_{y}(x,y) = b f^{'} (ax+by)[/tex],
[tex]u_{yy}(x,y) = b^2 f^{''} (ax+by)[/tex]
Finally, the solution of PDE is
[tex]u_{xx}(x,y) u_{yy}(x,y) - u_{xy} ^2 (x,y)[/tex]
[tex]= (a^2 f^{''} (ax+by)) (b^2 f^{''} (ax+by)) - (ab)^2 (f^{''} (ax+by))^2[/tex]
[tex]= (ab)^2 (f^{''} (ax+by))^2) - (ab)^2 (f^{''} (ax+by))^2[/tex]
= 0,
As the PDE is equal to 0, it means the function u(x,y) = f(ax+by) is the solution of the given PDE.
