English: And, we have, We have the initial conditions u(x,0) = f(x),   u_(t)(x,0) = g(x).Now, we can say from the given conditions that, ${\underline{u}}_{s}(\alpha,0) = F(\alpha),\ \ \ \frac{d{\underline{u}}_{s}(\alpha,0)}{dt} = G(\alpha)\ \ where\ \ F(\alpha) = F\left\lbrack f(x) \right\rbrack,\ \ \ G(\alpha) = F\lbrack g(x)\rbrack$ ∴ (3) implies A = F(α) and ${\underline{u}}_{s}(\alpha,t) = A\ cos\ c\alpha t + Bsin\ c\alpha t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$ ${\underline{u}}_{s}(\alpha,0) = F(\alpha),\ \ \ \frac{d{\underline{u}}_{s}(\alpha,0)}{dt} = G(\alpha)\ \ where\ \ F(\alpha) = F\left\lbrack f(x) \right\rbrack,\ \ \ G(\alpha) = F\lbrack g(x)\rbrack$ Therefore, we get, Now, taking the inverse Fourier sine transform, we have, $u(x,t) = F_{s}^{- 1}\left\lbrack {\underline{u}}_{s}(\alpha,t) \right\rbrack$ $= F_{s}^{- 1}\left\lbrack F(\alpha)cos\ c\alpha t\ + \frac{G(\alpha)}{c\alpha}sin\ c\alpha t\ \right\rbrack$ $= \left\lbrack F(\alpha)cos\ c\alpha t\ + \frac{G(\alpha)}{c\alpha}sin\ c\alpha t\ \right\rbrack sin\ \alpha x\ d\alpha$ $= \frac{1}{}F(\alpha)\left\lbrack sin\ \alpha(x + ct) + sin\ \alpha(x - ct)\ \ \right\rbrack d\alpha + \frac{1}{}\frac{G(\alpha)}{c\alpha}\left\lbrack cos\ \alpha(x - ct) - cos\ \alpha(x + ct)\ \ \right\rbrack d\alpha$ $u(x,t) = F_{s}^{- 1}\left\lbrack {\underline{u}}_{s}(\alpha,t) \right\rbrack$ $= F_{s}^{- 1}\left\lbrack F(\alpha)cos\ c\alpha t\ + \frac{G(\alpha)}{c\alpha}sin\ c\alpha t\ \right\rbrack$ $= \frac{1}{}F(\alpha)\left\lbrack sin\ \alpha(x + ct) + sin\ \alpha(x - ct)\ \ \right\rbrack d\alpha + \frac{1}{}\frac{G(\alpha)}{c\alpha}\left\lbrack cos\ \alpha(x - ct) - cos\ \alpha(x + ct)\ \ \right\rbrack d\alpha$ f(x) = F_(s)⁻¹[F(α)] ∴ f(x−ct) = F(α)sin α(x−ct) dαf(x+ct) Also, f(x) = F_(s)⁻¹[F(α)] ∴ f(x−ct) = F(α)sin α(x−ct) dαf(x+ct) Therefore, (5) implies, which is the required solution.
Telugu: మరియు, మాకు యు (x,0) = f(x), u_(t)(x,0) = g(x) అనే ప్రారంభ షరతులు ఉన్నాయి. ఇప్పుడు, ఇవ్వబడ్డ పరిస్థితుల నుండి, ${\underline{u}}{s}{s}(\alpha,0) = F(\\alpha),\ α ∴ \ \ 0) = F(\alpha),\ \ \frac{d{\underline{u}}{s}{s}(\alpha,0)}{dt} = G(\alpha)\ \\ F(\alpha) = F\లెఫ్ట్\lbrack f(x) \కుడి\rbrack,\ \ G(\alpha) = F\lbrack g(x)\rbrack$ అందువల్ల, ఇప్పుడు, మనం పొందుతాం, $u(x,t) = F_{s}^{1}\lbra F_ck {s}^{1}\లెఫ్ట్\lbrack {\underline{u}}{s}{s}(\alpha,t) ఆల్ఫా t\ \కుడి\rbrack$ $= \ఎడమ\lbrack F(\alpha)cos\ c\alpha t\ + \frac{G(\alpha)}{c\alpha}sin\ c\alpha t\ \right\rbrack sin\ \alpha x\ d\alpha$ $= \frac{1}{}F(\alpha)\లెఫ్ట్\lbrack sin\ \alpha(x + ct) + sin\ \alpha(x - ct)\ \alpha(x - ct)\ \\frac{1}{}\frac{g(\alpha)}{c\alpha}\left\lbrack cos\ \$u alpha(x + ct)\ \alpha(x + ct)\ \alpha(x + c\alpha)\ \frac{1}{frac{g(\alpha)}{c\alpha}{c\alpha}\table t) = F_{s}^{- 1}\లెఫ్ట్\lbrack {\underline{u}}{s}(\alpha,t) \కుడి\rbrack$ $= F_{s}^{- 1}\లెఫ్ట్\lbrack F(\alpha)cos\ c\alpha t\ + \frac{G(\alpha)}{c\alpha}sin\alpha t\ \right\rbrack$ $= \frac{1}{}F(\alpha)\fbrack sin\alpha(x + ct{1}) x) = F_(s)⁻¹[F(α)] ∴ f(x−ct) = F(α)sin α(x−ct) df(x+ct) అలాగే, f(x) = F_(s)⁻¹[F(α)] ∴ f(x−ct) = F(α)sin α(x−ct) dαf(x+ct)