11)[ordinary differential equations topic]Please solve the following problemProvide a well-explained and understandablestep by step solutionThank you A thin bar of length L= 3 meters is situated along the x axis so that one end is at x = 0 and the other end is at x=3. The thermal diffusivity of the bar is k = 0.4. The bar's initial temperature is given by the function f(x) = 100 - 25xdegrees Celsius. The ends of the bar (x = 0 and x = 3) are then put in an icy bath and kept at a constant 0 degreesC. Let u(x, t) be the temperature in the bar at x at time t, with t measured in seconds. Find u(x, t) and thenuz (2, 0.1).Put uz (2, 0.1) calculated accurately to the nearest thousandth (3 decimal places) in the answer box.

Given- a thin bar of length 3 meters such that one end is at x=0 and another end is at x=3. - The thermal diffusivity of the bar is k=0.4. -function for initial temperature f(x)=100-25x degree Celsius - The ends of the bar put in an icy bath and kept at 0 degrees C. to find- u(x,t) which is the temperature in the bar at x at time t. - and further u52,0.1the above question is solve using heat equation and heat equation for the given boundary condition u(0,t)=u(L,t)=0 is given by where Bn=2L∫0Lf(x)sinnπxLdx ; f(x)=u(x,0) u(x,t)=∑ n=1∞un(x,t)= ∑ n=1∞BnsinnπxLe-λn2t; λn=cnπL given thermal diffusivity k= c2=0.4 solution : here f(x)=100-25x ,L=3 therefore Bn=2L∫0Lf(x)sin(nπxL)dx =2L∫0L(100-25x)sin(nπxL)dx=50L∫0L(4-x)sin(nπxL)dx ==50L∫0L(4)sin(nπxL)dx -50L∫0L(x)sin(nπxL)dx=50L4Lnπ-cos(nπxL)0L - 50LxLnπ-cosnπxL+Lnπ∫cosnπxLdx0L=200nπ(1-(-1)n)-50LxLnπ-cosnπxL+Lnπ2sinnπxL0L=200nπ(1-(-1)n)+50nπL(-1)n=200nπ-200nπ((-1)n+150nπ(-1)nn)=200nπ-50nπ(-1)n thus u(x,t)=∑ n=1∞un(x,t)= ∑ n=1∞BnsinnπxLe-λn2t; λn=cnπL u(x,t)= ∑ n=1∞200nπ-50nπ(-1)nsinnπx3e-cnπL2t; λn=cnπLu(x,t)= ∑ n=1∞200nπ-50nπ(-1)nsinnπx3e-0.4n2π29t; u5(2,0.1)= 2005π-505π(-1)5sin2×5×π3e-0.4×52π290.1u5(2,0.1)= 200-50(-1)55πsin2×5×π3e-0.4×52π290.1u5(2,0.1)= 50π-320.33=7.672