English: Therefore, we obtain, $\ \Rightarrow \frac{d}{dt}\lbrack ucos\ \alpha x\ dx\rbrack = k\left\lbrack \left\lbrack \frac{\partial u}{\partial x}\cos\ \alpha x\ \right\rbrack_{0}^{\infty} + \alpha\frac{\partial u}{\partial x}\sin\ \alpha x\ dx \right\rbrack$ $\ \Rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\left\lbrack \mu + \alpha\left\{ \lbrack usin\ \alpha x\ \rbrack_{0}^{\infty} - \alpha ucos\ \alpha x\ dx \right\} \right\rbrack$ $\ \left\lbrack \because\frac{\partial u}{\partial x} \rightarrow 0\ \ as\ \ x \rightarrow \infty,\ \ u_{x}(0,t) = - \mu \right\rbrack$ $\ \Rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\mu - {k\alpha}^{2}ucos\ \alpha x\ dx\ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because u \rightarrow 0\ \ as\ \ x \rightarrow \infty\rbrack$ $F_{c}\left\lbrack \frac{\partial u}{\partial t} \right\rbrack = kF_{c}\left\lbrack \frac{\partial^{2}u}{\partial x^{2}} \right\rbrack$ $\ \Rightarrow \frac{d}{dt}\lbrack ucos\ \alpha x\ dx\rbrack = k\left\lbrack \left\lbrack \frac{\partial u}{\partial x}\cos\ \alpha x\ \right\rbrack_{0}^{\infty} + \alpha\frac{\partial u}{\partial x}\sin\ \alpha x\ dx \right\rbrack$ $\ \Rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\left\lbrack \mu + \alpha\left\{ \lbrack usin\ \alpha x\ \rbrack_{0}^{\infty} - \alpha ucos\ \alpha x\ dx \right\} \right\rbrack$ $\ \left\lbrack \because\frac{\partial u}{\partial x} \rightarrow 0\ \ as\ \ x \rightarrow \infty,\ \ u_{x}(0,t) = - \mu \right\rbrack$ $\ \Rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\mu - {k\alpha}^{2}ucos\ \alpha x\ dx\ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because u \rightarrow 0\ \ as\ \ x \rightarrow \infty\rbrack$ Thus, the given PDE is reduced to a first order ODE.
Telugu: అందువల్ల, మేము పొందుతాము, $\ \Rightarrow \frac{d}{dt}\lbrack ucos\ \alpha x\ dx\rbrack = k\left\lbrack \left\lbrack \frac{\పాక్షిక u}{\పాక్షిక x}\cos\ \alpha x\ \కుడి\rbrack_{0}^{\infty} + \alpha\frac{\పాక్షిక u}{\పాక్షిక x}\sin\ alpha x\ dx \కుడి\rbrack$ $\ \Rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\left\lbrack \mu + \alpha\left\{ \lbrack usin\ \alpha x rbrack_{0}\ u}{\పాక్షిక x} \rightarrow 0\ \ as\ x \rightarrow \infty,\ \ u_{x}(0,t) = - \mu \right\rbrack$ $\ \Rightarrow \frac{d{\underline{u}}{c}}{dt} = k\mu - {k\alpha}^{2}ucos\ \alpha x\ dx\ \\ \\ \ \ \ \\ \ \ \ \ \ \ \ \\ \row 0\ \ as\ \ x \rightarrow \infty\rbrack$ $F_{c}\left\lbrack \frac{\partial u}{\partial t} \కుడి\rbrack = kF_{c}\లెఫ్ట్\lbrack \frac{\partial^{2}u}{\పాక్షిక x^{2}} \కుడి\rbrack$ $\ \Rightarrow \frac{d}{dt}\lbrack ucos\ rbrack = k\left\lbrack \left\lbrack \frac{\పాక్షిక u}{\పాక్షిక x}\cos\ \alpha x\ \కుడి\rbrack_{0}^{\infty} + \alpha\frac{\పాక్షిక u}{\పాక్షిక x}\sin\ \alpha x\ dx \కుడి\rbrack$ $\ \rightarrow \frac{d{\underline{u}}_{c}{dt} = k\left\lbrack \mu + \alpha\left\{ \lbrack usin\ alpha x\ \rbrack_{0}^{\infty} - \alpha ucos\ \alpha x\ dx \కుడి\rbrack$ $\ \left\lbrack \ఎందుకంటే\frac{\పాక్షిక u}{\పాక్షిక x} \కుడిఅర్రో 0\ \agra\  \ u_{x}(0,t) = - \mu \rightarrow \rightarrow \frac{d{\underline{u}}_{c}}{dt} = k\mu - {k\alpha}^{2}ucos\ \alpha x\ dx\ \\ \lbrack\ఎందుకంటే u \rightarrow 0\ \ \ \\ \రైట్ ఆర్బ్రాక్\ \ as\ \x \rightarrow \infty\rbrack$