English: Therefore, we obtain, $F_{c}\left\lbrack \frac{\partial u}{\partial t} \right\rbrack = kF_{c}\left\lbrack \frac{\partial^{2}u}{\partial x^{2}} \right\rbrack$ $\Rightarrow \frac{d}{dt}\lbrack ucos\ \alpha x\ dx\rbrack = k\left\lbrack \left\lbrack \frac{\partial u}{\partial x}\cos\ \alpha x\ \right\rbrack_{0}^{\infty} + \alpha\frac{\partial u}{\partial x}\sin\ \alpha x\ dx \right\rbrack$ $\Rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\left\lbrack \mu + \alpha\left\{ \lbrack usin\ \alpha x\ \rbrack_{0}^{\infty} - \alpha ucos\ \alpha x\ dx \right\} \right\rbrack\left\lbrack \because\frac{\partial u}{\partial x} \rightarrow 0\ \ as\ \ x \rightarrow \infty,\ \ u_{x}(0,t) = - \mu \right\rbrack$ $\Rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\mu - {k\alpha}^{2}ucos\ \alpha x\ dx\ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because u \rightarrow 0\ \ as\ \ x \rightarrow \infty\rbrack$ $F_{c}\left\lbrack \frac{\partial u}{\partial t} \right\rbrack = kF_{c}\left\lbrack \frac{\partial^{2}u}{\partial x^{2}} \right\rbrack$ $\Rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\left\lbrack \mu + \alpha\left\{ \lbrack usin\ \alpha x\ \rbrack_{0}^{\infty} - \alpha ucos\ \alpha x\ dx \right\} \right\rbrack\left\lbrack \because\frac{\partial u}{\partial x} \rightarrow 0\ \ as\ \ x \rightarrow \infty,\ \ u_{x}(0,t) = - \mu \right\rbrack$ $\Rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\mu - {k\alpha}^{2}ucos\ \alpha x\ dx\ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because u \rightarrow 0\ \ as\ \ x \rightarrow \infty\rbrack$ Thus, the given PDE is reduced to a first order ODE.
Telugu: అందువల్ల, మేము $F_{c}\లెఫ్ట్\lbrack \frac{\పాక్షిక u}{\పాక్షిక t} \కుడి\rbrack = kF_{c}\లెఫ్ట్\lbrack \frac{\పాక్షిక^{2}u}{\partial x^{2}} \కుడి\rbrack$ $\Rightarrow \frac{d}{dt}\lbrack ucos\ \alpha x\ dx\rbrack = k\lbrack =k\lbrack \lbrack \lbrack \lbrack {\part rbrack_{0}\lbrack = k\lbrack \fbrack =k\lbrack \fbrack =k\lbrack \lbrack =k\lbrack \lbrack \lbrack = k\lbrack = k\lbrack \lbrack =k\lbrack \lbrack =k\lbrack = k\lbrack = k\lbrack =k\lbrack \lbrack = k\lbrack =k\lbrack \lbrack = k\lbrack =k\lbrack \lbrack =k\lbrack \lbrack = k\lbrack =k\lbrack \lbrack = k\lbrack =k\lbrack \lbrack = k\lbrack \lbrack = k\lbrack =k\lbrack \lbrack =k\l lbrack \mu + \alpha\left\{ \lbrack usin\ \alpha x\ \rbrack_{0}^{\infty} - \alpha ucos\ \alpha x\ dx \right\right\rbrack\లెఫ్ట్\lbrack \ఎందుకంటే\frac{\పాక్షిక u}{\పాక్షిక x} \rightarrow 0\ \\ \\కుడిఅర్రో \infty,\ \ u_{x}(0,t) = - \mu \rbrack$ $\Rightarrow \frac{d{\underline{u}}{c}}{c}{dt} = k\mu - {k\alpha}^{2}ucos\ \alpha x\ dx\ \lbrack\ఎందుకంటే u \rightarrow 0\ \ \rbrack \infty\rbrack $F$ కుడి\rbrack = kF_{c}\ఎడమ\lbrack \frac{\పాక్షిక^{2}u}{\పాక్షిక x^{2}} \కుడి\rbrack$ $\Rightarrow \frac{d{\underline{u}}{c}}{dt} = k\లెఫ్ట్\lbrack \mu + \alpha\లెఫ్ట్\{ \lbrack usin\ \alpha x\ \rbrack_{0}^{\infty} - \alpha ucos\ \alpha x\ dx \కుడి\rbrack\ఎడమ\lbrack \ఎందుకంటే\frac{\partial u}{\పాక్షిక x} \rightarrow 0\ \ \కుడిఅర్రో \in\ \  \ u_{x}(0,t) = - \mu \rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\mu - {k\alpha}^{2}ucos\ \alpha x\ dx\ \\ \lbrack\ఎందుకంటే u \rightarrow 0\ \ \ as\ \x \rightarrow \infty\rbrack$ అందువలన, ఇవ్వబడ్డ PDEని మొదటి ఆర్డర్ ODEకు తగ్గించారు.