A couple of comments. First, you say there's no solution for . Why can't always? (although it does lead to a trivial solution). Second, when you integrate the Chapit equation you should obtain arbitrary functions of and when then
Use Charpits Equations to Solve the following given that u=1 on the line x+2y=2.
Is the solution unique?
Initial Condition are
Therefore
Therefore
2 solutions p=0, p=1
No solution for p=0, therefore continuing with p=1 at t=0 to find A
.
Therefore . Now q=2p therefore .
Since 4up=q^2 implies u is
How am I doing so far?
The question asks Is the solution unique?
I am guessing that it is not unique because it has 2 valid solutions ie 1) p=0 and 2) p=1?
However, Im confused because when solving for p using given the condition that p=0 at t=0 I get
A(s)=0 which is not possible since A is constant, right?