- #1
GarageDweller
- 104
- 0
I thought of a method to solving poisons equation the other day, tell me if this is right.
we have, ∂(∂ψ/∂x)/∂x+∂(∂ψ/∂y)/∂y=P(x)A(y)
Assuming the solution is the product of two single variable functions,
G''(x)V(y)+V''(y)G(x)=P(x)A(y)
However, for the left hand side to equal the right hand side, we must be able to factor out either P(x) or A(y) from both terms, for the sake of this explanation, we will take P(x) to be the one that could be factored out, dividing by P(x) on both sides..
(G''(x)/P(x))V(y)+(G(x)/P(x))V''(y)=A(y)
Now from this, we can see that G''(x)/P(x) and G(x)/P(x) equal some pair of constants (k,q)
kV(y)+qV''(y)=A(y)
G''(x)+G(x)=(k+q)P(x)
we have, ∂(∂ψ/∂x)/∂x+∂(∂ψ/∂y)/∂y=P(x)A(y)
Assuming the solution is the product of two single variable functions,
G''(x)V(y)+V''(y)G(x)=P(x)A(y)
However, for the left hand side to equal the right hand side, we must be able to factor out either P(x) or A(y) from both terms, for the sake of this explanation, we will take P(x) to be the one that could be factored out, dividing by P(x) on both sides..
(G''(x)/P(x))V(y)+(G(x)/P(x))V''(y)=A(y)
Now from this, we can see that G''(x)/P(x) and G(x)/P(x) equal some pair of constants (k,q)
kV(y)+qV''(y)=A(y)
G''(x)+G(x)=(k+q)P(x)