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

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- Aug 19th 2013, 12:43 AMiamapineappleHow 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 AMiamapineappleRe: 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 AMibduttRe: How to solve dt/dT = -K(T - 5) + H for T?
- Aug 19th 2013, 03:02 AMProve ItRe: How to solve dt/dT = -K(T - 5) + H for T?
- Aug 19th 2013, 04:05 AMiamapineappleRe: 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 AMHallsofIvyRe: 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.