the problem is posed incorrectly since the initial conditions and boundary conditions contradict each other:
u(x,0)=sin(3x) and u(4,t)=0,but sin(3⋅4)=0 Suppose
u(x,0)=sin(3πx) We apply the Laplace transform to the function u(x,t) in t, considering x as a parameter
L{u(x,t)}=U(x,s)
L{dtdu(x,t)}=sU(x,s)−u(x,0)=sU(x,s)−sin(3πx)
L{dx2d2u(x,t)}=Uxx(x,s) Applying the Laplace transform to the equation, we have
Uxx(x,s)=sU(x,s)−sin(3πx) The general solution can be written as
U(x,s)=Uh(x,s)+Up(x,s) the general solution of the homogeneous equation
Uh(x,s)=C1exs+C2e−xs the particular solution of the non-homogeneous equation
Up(x,s)=C3cos(3πx)+C4sin(3πx)
dx2d2Up(x,s)=−(3π)2Up(x,s) substitute the particular solution into the equation
−(3π)2Up(x,s)=sUp(x,s)−sin(3πx)
(−(3π)2−s)(C3cos(3πx)+C4sin(3πx))=−sin(3πx)
(−(3π)2−s)C3=0 → C3=0
(−(3π)2−s)C4=−1 → C4=s+9π21 the general solution
U(x,s)=C1exs+C2e−xs+s+9π2sin(3πx) from the boundary condition
u(0,t)=0→U(0,s)=0=C1+C2+0
u(4,t)=0→U(4,s)=0=C1e4s+C2e−4s+0 then
C1=C2=0 So, the solution is
U(x,s)=s+9π2sin(3πx) apply the inverse Laplace transform
u(x,t)=L−1{s+9π2sin(3πx)}=sin(3πx)L−1{s+9π21}
u(x,t)=e−9π2tsin(3πx), t≥0 Answer: u(x,t)=e−9π2tsin(3πx), t≥0, 0≤x≤4
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