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

1. ## 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.

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

$T =$ $\frac{1}{k}$ $(Ae\textsuperscript{-kt} + 5k + H)$

Is this correct?

3. ## 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?

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

Is this correct?

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

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.

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

Shoot, sorry, the equation was dT/dt = the equation

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

IF the problem was $\frac{dt}{dT}= -K(T- 5)+ H$ then write it as $dt= (-K(T- 5)+ H)dT$ and integrate both sides.

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