dh/dt = (6-h)/20. When t=0, h=1. Show that t=20ln(5/(6-h))

Part 1: This question requires us to first do some rearranging. The right hand side of the equation is in terms of h, and we are looking to find t in terms of h, so it won't be appropriate to simply integrate both sides with respect to t. Therefore we need to rearrange to separate the variables. We can do this by first dividing both sides by (6-h), and then integrating both sides with respect to t (you can think of it as 'multiplying by dt'). We then have ∫1/(6-h) dh = ∫1/20 dt.Part 2: IntegratingLHS: Use the reverse chain rule. You know that ln(6-h)/dh= =-1/(6-h) by the chain rule i.e. ∫1/(6-h) dh = -ln(6-h)+ a so we just need to account for this multiplying by -1. Therefore ∫1/(6-h) dh = -ln(6-h)+aRHS: 1/20 is just a constant so you get ∫1/20 dt = t/20+bSo -ln(6-h)=t/20+c (where c is just b-a)Part 3: Sub in initial conditions If we sub in t=0 and h=1, we get -ln5=cSo -ln(6-h)=t/20-ln5. By using laws of logs this gives t/20=ln(5/(6-h)), and therefore t=20ln(5/6-h).