Question

How do you find the particular solution to `dT+k(T-70)dt=0` that satisfies `T(0)=140`?

Answer 1

`T(t) = 70 + 70e^(-kt)`

Explanation:

We have `dT + k(T-70)dt = 0` which is a First Order separable DE

We can rearrange as follows:

`dT =- k(T-70)dt => (dT)/(k(T-70)) = -dt`
`:. 1/kint 1/(T-70) dT= - int dt`

Integrating gives:
`:. 1/kln (T-70) = -t+C`

I always prefer to aptly the initial conditions as soon as possible to minimise the change of an algebraic slip:

`T=140` when `t=0`
`:. 1/kln (140-70) = C => 1/kln (70)`

Substituting back into our DE solution gives us:
`1/kln (T-70) = -t+1/kln (70)`
`:. ln (T-70) = -kt+ln (70)`
`:. T-70 = e^(-kt+ln (70))`
`:. T-70 = e^(-kt)e^(ln (70))`
`:. T-70 = e^(-kt)(70)`
`:. T = 70 + 70e^(-kt)`