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

Nov 8, 2016

$T \left(t\right) = 70 + 70 {e}^{- k t}$

#### Explanation:

We have $\mathrm{dT} + k \left(T - 70\right) \mathrm{dt} = 0$ which is a First Order separable DE

We can rearrange as follows:

$\mathrm{dT} = - k \left(T - 70\right) \mathrm{dt} \implies \frac{\mathrm{dT}}{k \left(T - 70\right)} = - \mathrm{dt}$
$\therefore \frac{1}{k} \int \frac{1}{T - 70} \mathrm{dT} = - \int \mathrm{dt}$

Integrating gives:
$\therefore \frac{1}{k} \ln \left(T - 70\right) = - t + C$

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

$T = 140$ when $t = 0$
$\therefore \frac{1}{k} \ln \left(140 - 70\right) = C \implies \frac{1}{k} \ln \left(70\right)$

Substituting back into our DE solution gives us:
$\frac{1}{k} \ln \left(T - 70\right) = - t + \frac{1}{k} \ln \left(70\right)$
$\therefore \ln \left(T - 70\right) = - k t + \ln \left(70\right)$
$\therefore T - 70 = {e}^{- k t + \ln \left(70\right)}$
$\therefore T - 70 = {e}^{- k t} {e}^{\ln \left(70\right)}$
$\therefore T - 70 = {e}^{- k t} \left(70\right)$
$\therefore T = 70 + 70 {e}^{- k t}$