Givenpde:∂t∂u=2∂x2∂2u+x+2,0<x<1,t>0u(x,0)=x+2,0<x<1u(0,t)=2u(2,t)=−2The given pde is non homogeneous heat equation.Now, let u(x,t)=v(x,t)+Ax+BTherefore, ∂t∂u=∂t∂v, ∂x2∂2u=∂x2∂2vWe get, A=B=2.Then,v(x,0)=x,0<x<1v(0,t)=0v(2,t)=0Therefore, the solution is given byv(x,t)=∑n=1∞En(x)sin(lnπx)el2−αn2π2t+∑n=1∞sin(lnπx)∫0tel2−αn2π2(t−s)Fn(s)dsWhere En(x)=l2∫0lF(x)sin(lnπx)dx,Fn(s)=l2∫0tϕ(s,t)sin(lnπs)dsHere, we haveF(x)=x,ϕ(x,t)=x+2,α=2,l=2.Therefore, En(x)=∫02xsin(2nπx)dx=nπ(−1)n+14Fn(s)=∫0∞(s+2)sin(2nπs)dsTherefore, solution isu(x,t)=∑n=1∞En(x)sin(lnπx)el2−αn2π2t+∑n=1∞sin(lnπx)∫0tel2−αn2π2(t−s)Fn(s)ds+2x+2where En(x)=nπ(−1)n+14Fn(s)=∫0∞(s+2)sin(2nπs)ds
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