[ordinary differential equations topic]Please solve the following problemProvide a well-explained and understandablestep by step solution Thank you e) 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) = 180 - 20xdegrees 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 thenu4 (2, 0.1).Put us (2, 0.1) calculated accurately to the nearest thousandth (3 decimal places) in the answer box.

Given that fx=180-20x PDE is ut=Kuxx x∈0,L,K≥0L=3,k=0.7 BC is u0,t=uL,t=0 IC is ux,0=fx=180-2x. The general solution is ux,t=∑n=1∞BnsinnπxLe-n2π2ktL2 →1fx=A02+∑n=1∞AncosnπxL+∑n=1∞BnsinnπxL ,n=0,1,2,3,... →2An=1L∫-LLf(x)cosnπxLdx Bn=1L∫-LLf(x)sinnπxLdx n=1,2,3,...ux,0=fxBnsinnπxL →3 From 1 and 2 find Bn using the standard formula Bn=2L∫0Lf(x)sinnπxLdx =2L∫0L(180-20x)sinnπxLdxBn=2L180∫0LsinnπxLdx-20∫0LxsinnπxLdxNow, to find the integral values of Bn ∫0LsinnπxLdx=-LnπcosnπxL0L =-Lnπcosnπ-cos 0 = -Lnπ-1n-1 ∫0LsinnπxLdx=Lnπ1--1n ∫0LxsinnπxLdx=x-LnπcosnπxL0L-∫0L-LnπcosnπxLdx =L-Lnπcosnπ+Lnπ∫0LcosnπxLdx =-L2nπ-1n+LnπLnπsinnπxL0L =-L2nπ-1n+0∫0LxsinnπxLdx=L2nπ-1n+1 Bn=2L180Lnπ1--1n-20L2nπ-1n+1Bn=360nπ1--1n-40Lnπ-1n+1Bn=360nπ1--1n-120nπ-1n+1 ,L=3From 3 ux,t=∑n=1∞unx,t =∑n=1∞360nπ1--1n-120nπ-1n+1sinnπx3e-0.4n2π2t9u42,0.1=3604π1-1-1204π-14+1sin4π23e-0.442π20.19u42,0.1=9.554140.86600.496=4.103 Thus the solution is u42,0.1=4.103