Re: Newton's law of cooling

$\displaystyle \frac{dT}{dt}+ 0.2T- 5= 0$ is the same as $\displaystyle \frac{dT}{dt}= 5- 0.2T$.

Changing to differential form $\displaystyle \frac{dT}{5- 0.2T}= dt$. Can you integrate those?

Re: Newton's law of cooling

Quote:

Originally Posted by

**leonm92**

**separation of variables ...**

$\displaystyle \frac{dT}{dt} = 5 - 0.2T$

$\displaystyle \frac{dT}{dt} = 0.2(25 - T)$

$\displaystyle \int \frac{dT}{25-T} = \int 0.2 \, dt$

finish it and get temperature $\displaystyle T$ as a function of time $\displaystyle t$ ...

Re: Newton's law of cooling

to be honest, I haven't got a clue about integrating

Re: Newton's law of cooling

Quote:

Originally Posted by

**leonm92** to be honest, I haven't got a clue about integrating

then it's obvious you are ill prepared to tackle this problem ... here is a link for a place to start

Indefinite Integrals Part 1 | Khan Academy Calculus Lecture

Re: Newton's law of cooling

Quote:

Originally Posted by

**leonm92** to be honest, I haven't got a clue about integrating

It is not a good idea to take differential equations until after you have learned calculus.