I have a question about "the two-dimensional wave equation", I have solved it but I wanted to know if I've done right. Because the solution is long, I just write the answer but I can write the solution also if you need it. Thanks in advance
Solve two-dimensional wave equation $$\frac{∂^2u}{∂t^2}=4*(\frac{∂^2u}{∂x^2}+\frac{∂^2u}{∂y^2})$$ in $$0<x<3$$$$0<y<2$$
then we have boundary conditions $$\frac{∂u}{∂x}(0,y,t)=0 $$$$\frac{∂u}{∂x}(2,y,t)=0$$$$\frac{∂u}{∂y}(x,0,t)=0$$$$u(x,2,t)=0$$ and we have and the initial conditions $$u(x,y,0)=0$$$$\frac{∂u}{∂t}(x,y,0)=g(x,y)$$
My answer is: $$u(x,y,t)=\sum\sum A_ncos(\frac{pi}{3(1+n)x})*A_mcos(\frac{pi}{2(m+1/2)y})*A_(nm)cos(2 \sqrt{λ})t B_(nm)sin(2 \sqrt{λ})t)$$ but we have $$A_(nm)=0$$ so $$u(x,y,t)=\sum\sum A_ncos(\frac{pi}{3(1+n)x})*A_mcos(\frac{pi}{2(m+1/2)y})*[B_(nm)sin(2sqrt λ)t)$$ and $$b_(nm)=\frac{<f,Z_(nm)>}{<Z_(nm),Z_(nm)>}=$$ $$=\frac{\int_0^3\int_0^2g(x,y)cos(\frac{pi}{3(1+n)x})*cos(\frac{pi}{2(m+1/2)y})dydx}{2\sqrt{λ_{nm}}\int_0^3\int_0^2cos^2(\frac{pi}{3(1+n)x})cos^2(\frac{pi}{2(m+1/2)y})dydx}$$