$\frac{de{l}^{2}u}{del{x}^2}+\frac{de{l}^{2}u}{del{y}^2}+g_0=0in0xL,0y1$

$B.C.\text{'}s$: $\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{x}}=0atx=0,0y1$

$u=0atx=L,0y1$

$\frac{\text{d}\text{e}\text{l}\text{u}}{\text{d}\text{e}\text{l}\text{y}}=0aty=0,0x1$

$u=0aty=1,0x1$

Solve the problem with the assumption $of$

$\frac{?}{\text{b}\text{a}\text{r}}(E)q*u(x,y)\frac{?}{b}=ar(v)(x,y)\frac{+}{b}ar(w)(x)$

Will this solution $be$ different from the assumption

$u(x,y)=v(x,y)+w(y),$ why?

show these forms are esentially equivalent $to$ each other.