Thanks I get
c.ln(x) + C=ln(X) => X=Ax^C
c.ln(y) +C= 1/2 ln(Y) => Ay^2C
so u(x,y)=X.Y=A^2(y^2C)(x^C)
...the contants for integration are required right? Just wondering because the second part of the question would be easier if they weren't there:
determine u(x,y) which satisfies the...
Find the general solution of:
x.du/dx - (1/2).y.du/dy=0
I know that for it to be separable u(x,y)=X(x)Y(y)
so:
x.Y(y)dX/dx + (1/2).X(x)dY/dy = 0
which cancels to:
x/X(x).(dX/dx) = - y/2Y(y).dY/dy
so:
X(x) = -c. Y(y) c is some constant
so:
x/X(x).dX/dx = c...