# How to solve dt/dT = -K(T - 5) + H for T?

• Aug 19th 2013, 12:43 AM
iamapineapple
How to solve dt/dT = -K(T - 5) + H for T?
After flipping it to get dT/dt, how would I solve the integral and rearrange for T? Help appreciated.
• Aug 19th 2013, 01:03 AM
iamapineapple
Re: How to solve dt/dT = -K(T - 5) + H for T?
$\displaystyle T = $$\displaystyle \frac{1}{k}$$\displaystyle (Ae\textsuperscript{-kt} + 5k + H)$

Is this correct?
• Aug 19th 2013, 02:03 AM
ibdutt
Re: How to solve dt/dT = -K(T - 5) + H for T?
How we got this. As i understand we get an equation like:
dt = [ -K(T-5) + H ] dT
Now integrating we get
(-K+H) T^2 + 10 T = t
We have to then solve this quadratic for T.
Is that correct?

Quote:

Originally Posted by iamapineapple
$\displaystyle T = $$\displaystyle \frac{1}{k}$$\displaystyle (Ae\textsuperscript{-kt} + 5k + H)$

Is this correct?

• Aug 19th 2013, 03:02 AM
Prove It
Re: How to solve dt/dT = -K(T - 5) + H for T?
Quote:

Originally Posted by iamapineapple
After flipping it to get dT/dt, how would I solve the integral and rearrange for T? Help appreciated.

You don't flip the integral at all as your function is already in terms of T. Just integrate both sides.
• Aug 19th 2013, 04:05 AM
iamapineapple
Re: How to solve dt/dT = -K(T - 5) + H for T?
Shoot, sorry, the equation was dT/dt = the equation
• Aug 19th 2013, 06:12 AM
HallsofIvy
Re: How to solve dt/dT = -K(T - 5) + H for T?
IF the problem was $\displaystyle \frac{dt}{dT}= -K(T- 5)+ H$ then write it as $\displaystyle dt= (-K(T- 5)+ H)dT$ and integrate both sides.

IF it was $\displaystyle \frac{dT}{dt}= -K(T- 5)+ H$ then write it as $\displaystyle \frac{dT}{(T- 5- H/K}= -Kdt$ and integrate both sides.
(To integrate on the left, let u= T- 5- H/K.) T will be an exponential of t.