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?

$\displaystyle T = $$\displaystyle \frac{1}{k}$$\displaystyle (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
$\displaystyle T = $$\displaystyle \frac{1}{k}$$\displaystyle (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 $\displaystyle \frac{dt}{dT}= -K(T- 5)+ H$ 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.