English: $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow {\underline{u}}_{c}(\alpha,t) = A{\ e}^{- k\alpha^{2}t} + \ \frac{\mu}{\alpha^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$ Now, we are provided with the following initial condition that which implies that Therefore, (1) implies $0 = A + \ \frac{\mu}{\alpha^{2}} \Rightarrow A = - \ \frac{\mu}{\alpha^{2}}$ $0 = A + \ \frac{\mu}{\alpha^{2}} \Rightarrow A = - \ \frac{\mu}{\alpha^{2}}$ Hence, the solution is obtained from (1) as $\Rightarrow {\underline{u}}_{c}(\alpha,t) = \ \frac{\mu}{\alpha^{2}}\left( 1 - {\ e}^{- k\alpha^{2}t} \right)$ $\Rightarrow {\underline{u}}_{c}(\alpha,t) = \ \frac{\mu}{\alpha^{2}}\left( 1 - {\ e}^{- k\alpha^{2}t} \right)$ Now, taking the inverse Fourier cosine transform, we have, $u(x,t) = F_{c}^{- 1}\left\lbrack {\underline{u}}_{c}(\alpha,t) \right\rbrack$ $= \mu\ F_{c}^{- 1}\left\lbrack \frac{1}{\alpha^{2}}\left( 1 - {\ e}^{- k\alpha^{2}t} \right) \right\rbrack$ Evaluation of the above integral will give the required solution for $u(x,t) = F_{c}^{- 1}\left\lbrack {\underline{u}}_{c}(\alpha,t) \right\rbrack$ $= \mu\ F_{c}^{- 1}\left\lbrack \frac{1}{\alpha^{2}}\left( 1 - {\ e}^{- k\alpha^{2}t} \right) \right\rbrack$ Now, we want to solve the next problem using Fourier sine transform: $\ u,\ \frac{\partial u}{\partial x} \rightarrow 0\ \ as\ \ x \rightarrow \infty$ $\ u,\ \frac{\partial u}{\partial x} \rightarrow 0\ \ as\ \ x \rightarrow \infty$ We apply Fourier sine transform with respect to x on the given equation.
Telugu: $\ \){{2}} =A{\ e}^{k\alpha^{2} {2}t} + \frac{\mu {2}}{\frac{\{2}} \ \){{2}} \రైట్రో ఎ = - \\frac {2}{u}{\){\alpha}{\alpha^{2}}  పరిష్కారం (1) నుండి $\rightarrow {\inline{u}}{c}(\alpha,t) = \\frac{\mu}{\alpha^{2}}\లెఫ్ట్( 1 - {\ e}^{- k\alpha^{2}t} \$u కుడి)$ $\Rightarrow {\underline{u}}{c}(\alpha,t) = \\frac{\mu}{\alpha^F_ {2}}\లెఫ్ట్, {\e}^{k\alpha^{2}t} t) \కుడి\rbrack$ $= \mu\ F_{c}^{- 1}\లెఫ్ట్\lbrack \frac{1}{\alpha^{2}}\ఎడమ( 1 - {\ e}^{- k\alpha^{2}t} \కుడి) \కుడి\rbrack$ పై అంతర్భాగం యొక్క మూల్యాంకనం $u(x,t) = F_{c}^{- 1}\ఎడమ\lbrack {\F_{c}{1}{1}{{2}}\లెఫ్ట్\lbrack {\inline{u}}{c}{c}{1}\lbrack \frac{1}{1}{alpha}}{c}{1}\ఎడమ\lbrack {1- {\ e}^{k\alpha^{2}t} \కుడి) \కుడి\rbrack$ Now,  మేము ఫోరియర్ సైన్ రూపాంతరాన్ని ఉపయోగించి తదుపరి సమస్యను పరిష్కరించాలనుకుంటున్నాము: $\ u,\ \frac{\పాక్షిక u}{\పాక్షిక x} \rightarro 0\ \as\ \x \rightarro \infty$ $\ u,\ \frac{\పాక్షిక u}{\పాక్షిక x} \rightarro 0\ \as\ \x \కుడిఅర్రో \infty$ ఇవ్వబడ్డ సమీకరణంలో xకు సంబంధించి ఫోరియర్ సైన్ రూపాంతరాన్ని వర్తింపజేస్తాము.