English: $\therefore u(x,t) = \frac{1}{2}\left\lbrack \frac{1}{}\left\{ \frac{1}{}f(u)e^{- isu}du \right\}\left( e^{- icst} + e^{icst} \right)e^{isx}ds \right\rbrack$ $\ \ \ \ \ \ \ \ \ \ \ \ = \ \ \frac{1}{2}\left\lbrack \frac{1}{}\left\{ \frac{1}{}f(u)e^{- isu}du \right\}\left\{ e^{is(x - ct)} + e^{is(x + ct)} \right\} ds \right\rbrack$ $\ \ \ \ \ \ \ \ \ = \ \frac{1}{2}\left\lbrack \frac{1}{}e^{is(x - ct)}\left\{ \frac{1}{}f(u)e^{- isu}du \right\} ds + \frac{1}{}e^{is(x + ct)}\left\{ \frac{1}{}f(u)e^{- isu}du \right\} ds \right\rbrack$ $= \frac{1}{2}\left\lbrack f(x - ct) + f(x + ct) \right\rbrack$ In deriving the above, we have used the well-known property of Dirac and that Dirac delta function is an even function i.e., $\therefore u(x,t) = \frac{1}{}\left\lbrack \frac{1}{}f(u)e^{i\alpha u}du \right\rbrack\left\lbrack \frac{e^{ic\alpha t} + e^{- ic\alpha t}}{2} \right\rbrack e^{- i\alpha x}d\alpha$ $\therefore u(x,t) = \frac{1}{2}\left\lbrack \frac{1}{}\left\{ \frac{1}{}f(u)e^{- isu}du \right\}\left( e^{- icst} + e^{icst} \right)e^{isx}ds \right\rbrack$ $\ \ \ \ \ \ \ \ \ \ \ \ = \ \ \frac{1}{2}\left\lbrack \frac{1}{}\left\{ \frac{1}{}f(u)e^{- isu}du \right\}\left\{ e^{is(x - ct)} + e^{is(x + ct)} \right\} ds \right\rbrack$ $= \frac{1}{2}\left\lbrack f(x - ct) + f(x + ct) \right\rbrack$ Therefore, the obtained solution $u(x,t) = \frac{1}{2}\left\lbrack f(x - ct) + f(x + ct) \right\rbrack$ is known as the D’Alembert’s solution of wave equation.
Telugu: $\అందువల్ల u(x,t) = \frac{1}{2}\లెఫ్ట్\lbrack \frac{1}{}\లెఫ్ట్\{ \frac{1}{}f(u)e^{- isu}du \కుడి\}\ఎడమ( e^{- icst} + e^{icst} \కుడి)e^{isx}ds \కుడి\rbrack$ $\ \ \frac{1}{2} \\ \ \\frac{1}{2} \\frac{1}{1}{2} \\frac{1}{2}{}\ఎడమ\{ \frac{1}{}f(u)e^{- isu}du \కుడి\}\లెఫ్ట్\{ e^{is(x - ct)} + e^{is(x + ct)} \కుడి\} ds \కుడి\rbrack$ $\ \{1}{2} \lbrack \lbrack \frac{1}{}e^{is(x - ct)}\frac{1}{}f( u)e^{- isu}du \కుడి\} ds + \frac{1}{}e^{is(x + ct)}\లెఫ్ట్\{ \frac{1}{}f(u)e^{- isu}du \కుడి\} ds \కుడి\rbrack$ $= \frac{1}{2}\లెఫ్ట్\lbrack f(x - ct) + f(x + ct) \కుడి\rbrack$ పైవాటిని పొందడంలో, మనం డిరాక్ యొక్క సుప్రసిద్ధ లక్షణాన్ని ఉపయోగించాము.  $\అందువల్ల u(x,t) = \frac{1}{}\left\lbrack \frac{1}{}f(u)e^{i\alpha u}du \right\rbrack\left\lbrack \frac{e^{ic\alpha t} + e^{- ic\alpha t}}{2} \కుడి\rbrack e^{- i\alpha x}d\alpha$ $\అందువల్ల u(x,t) = \frac{1}{2}\left\lbrack \frac{1}{}\left\{ \frac{1}{}f(u)e^{- isu}du \right\}\it( e^{{1}{2}icst} frac{1}{}\ఎడమ\{ {frac{1}{}f(u)e^{- isu}du \కుడి\}\ఎడమ\{ e^{is(x - ct)} + e^{is(x + ct)} \కుడి\} ds \కుడి\rbrack$ $= \frac{1}{2}\లెఫ్ట్\lbrack f(x - ct) + f(x + ct) \కుడి\rbrack$ అందువల్ల, పొందిన పరిష్కారం $u(x,t) = \frac{1}{2}\lbrack f(x - ct)