Please helpSolutiona d2Tdx2 hTa T 0 we are given that h

Please help!

Solution

a>

d^2T/dx^2 + h(Ta - T) = 0

we are given that h = .02/m^2 and Ta = 20 degree celcius
and the boundary conditions T(x=0) = 40 degree celcius and T(x=10) = 200 degree celcius

d^2T/dx^2 = .02(20-T)

the general solution of these kind of equation is of the form:

T(x) =Asin(rx) + Bcos(rx)

A and B are the constant of integrtion which we can find using the boundary conditions

r = .141421

=> T(x) =Asin(.141421x) + Bcos(.141421x) + 20
now at x = 0 ; T = 40
and at x=10 ; T = 200

=> A = 179.072 and B = 20

T(x) =179.072*sin(.141421x) + 20*cos(.141421x) + 20