Consider the following partial differential equation: y(e +1)u, – e"(y² + 1)²u, = 0, y >0 1. By using the separation method with u(r, y) = f(r)+ g(y) to solve (F), we obtain that : (F) 1+ e" df (y² +...

0 1. By using the separation method with u(r, y) = f(r)+ g(y) to solve (F), we obtain that : (F) 1+ e" df (y² + 1)² dg a) da fip df b) 1+ er dr (y? + 1)? dg dy dg et %3D 1+ e" df c) et dr (y? +1)2 dy d) None of the above. 2. A solution u(x, y) of (F) is ( B is an arbitrary constant): a) u(r, y) = À (In(1+ e*) – In(y² +1)) + B. 1 b) u(x, y) = (In(1 + e")- + B. 2(y² + 1) 1 c) u(r, y) = A + B. 2(y? + 1). d) None af the above. "/>
Extracted text: Consider the following partial differential equation: y(e +1)u, – e"(y² + 1)²u, = 0, y >0 1. By using the separation method with u(r, y) = f(r)+ g(y) to solve (F), we obtain that : (F) 1+ e" df (y² + 1)² dg a) da fip df b) 1+ er dr (y? + 1)? dg dy dg et %3D 1+ e" df c) et dr (y? +1)2 dy d) None of the above. 2. A solution u(x, y) of (F) is ( B is an arbitrary constant): a) u(r, y) = À (In(1+ e*) – In(y² +1)) + B. 1 b) u(x, y) = (In(1 + e")- + B. 2(y² + 1) 1 c) u(r, y) = A + B. 2(y? + 1). d) None af the above.