Show that T(x) = TL + (T0 - TL) * sin h(k(L - x))/sin h(kL) is a solution to the
ID: 1888807 • Letter: S
Question
Show that T(x) = TL + (T0 - TL) * sin h(k(L - x))/sin h(kL) is a solution to the 2nd order DE: d2T/dx2 = -k2(TL - T)Explanation / Answer
T(x) = TL + (T0-TL) sin h(k(L-x))/ sin h(kL) => dT/dx = dTL/dx + (T0-TL) (-k)[cos h(k(L-x))/sin h(kL)] [by differentiating both sides] dT/dx = (T0-TL)(-k) cos h(k(L-x))/sin h(kL) [since TL is constant wrt x, dTL/dx = 0] d^2T/dx^2 = (T0-TL)(k^2) sin h(k(L-x))/sin h(k(L-x) [differentiating again both sides] --(1) From original equation, we have (T(x) - TL)/(T0-TL) = sin h (k(L-x))/sin h(kL) put in the obtained eq. (1) d^2T/dx^2 = (T0-TL) * (k^2) * [(T(x)-TL)/(T0-TL)]= k^2(T-TL) = -k^2(TL-T) since d^2T/dx^2 = -k^2(TL-T) which is the given 2nd Order DE, Hence proved that T is the solution of d^2T/dx^2 = -k^2(TL-T)
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