∂T∂u=8∂x2∂2uu(0,t)=u(5,t)=0u(x,0)=2sinπx−4sin2πx Let's find the solution in the form
u(x,t)=X(x)T(t)XT′=8X′′TTT′(t)=8XX′′(x)=−8λ=const where lambda is an unknown constant.
From the boundary conditions
X′′+λX=0X(0)=X(5)=0X(x)=Csin5nπxλ=(5πn)2 where n is integer.
Hence
T′=−8λT→T(t)=Cexp(−8λt) The heat equation is linear, so the solution can be written in the form
u(x,t)=n=1∑∞CnXn(x)Tn(t) where Cn are constants determined from the initial conditions and Xn, Tn are the functions found above with the integer parameter n (for n = 0 X0(x)= 0).
From the initial conditions
u(x,t)=2exp(−852π252t)sinπx−4(−852π2102t)sin2πxu(x,t)=2e−8π2tsinπx−4e−32π2tsin2πx
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