Use the Laplace transform to solve $U_t=kU_{xx} $ in$ (0,l)$ with $U_{x}(0,t)=0$, $U_{x}(l,t)=0$ and $U(x,0)=1+cos({2 \pi x \over l})$
.The answer I get is $U_{x,t}=cos({2 \pi x \over l}) e^{-k4\pi^2t/l^2}$ But the given answer in the book is
$U_{x,t}=1+cos({2 \pi x \over l}) e^{-k4\pi^2t/l}$
This is what I did.Can someone please point out where I am making a mistake.
After taking Laplace transform on both sides I get $kU_{xx}=sU(x,s)=-[{1+cos({2 \pi x \over l})] }$.
Then the complementary solution is $C_1e^{\sqrt{ s\over k}x}+C_2e^{-\sqrt{ s\over k}x} $.
To find the particular solution I let $Y=A cos({2 \pi x \over l})+B sin({2 \pi x \over l})$
Then $-k({{2\pi \over l}})^2[A cos({2 \pi x \over l})+B sin({2 \pi x \over l}])-s[A cos({2 \pi x \over l})+B sin({2 \pi x \over l})]=-[1+A cos({2 \pi x \over l})]$
Then I compared coefficients of $cos({2 \pi x \over l})$ and $sin({2 \pi x \over l})$.When the constant term is compared 1=0.Therefore should I add a constant term for the left hand side.Is that how the answer has a +1.What's wrong with what I have done.
In your particular solution, a constant term is missing which amounts to missing $1$ in your answer. In particular solution, you could do without a sine term, but not without the constant term. You need to find $Y=A\cos{\bigl(\frac{2\pi x}{l}\bigr)}+B\,$ satisfying the equation $\,kY_{xx}=sY-\bigl[1+\cos{\bigl(\frac{2\pi x}{l}\bigr)}\bigr]$, which implies that $B=1/s$ while $A$ stays the same.