Answer:-
Laplace transform:
L(u(x,t))=∫0∞e−stu(x,t)dt=U(x,s)
L(ut(x,t))=∫0∞e−stut(x,t)dt=e−stu(x,t)∣0∞+s∫0∞e−stu(x,t)dt=
=sU(x,s)−U(x,0)
L(ux(x,t))=∫0∞e−stux(x,t)dt=Ux(x,s)
Then:
sU(x,s)−U(x,0)=U(x,s)−Ux(x,s)
dxdU(x,s)+(s−1)U(x,s)=e−7x
integrating factor:
μ=e∫(s−1)dx=e(s−1)x
Thus we have:
dxd[e(s−1)xU(x,s)]=e(s−1)xe−7x
U(x,s)=e−(s−1)x∫0xe(s−1)re−7rdr+ce−(s−1)x=
=e−(s−1)x⋅s−8e(s−8)r∣0x+ce−(s−1)x=s−8e−7x−e−(s−1)x+ce−(s−1)x
Taking the inverse Laplace transform we have:
u(x,t)=e−7xe8t−ex+8(t−x)H(t−x)+cetδ(t−x)
where H is Heaviside function:
H(x)={01x<0x≥0
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