Given the differential equation : `dT+K(T-70)dt=0, T(0)=140`

We have to find a particular solution that satisfies the initial condition.

We can write,

`dT=-K(T-70)dt`

`\frac{dT}{T-70}=-Kdt`

`\int \frac{dT}{T-70}=\int -Kdt`

`ln(T-70)=-Kt+C` where C is a constant.

Now,

`T-70=e^{-Kt+C}`

`=e^{-Kt}.e^{C}`

`=C'e^{-Kt}` where `e^C=C'` is again a constant.

Hence we have,

`T=70+C'e^{-Kt}`

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Given the differential equation : `dT+K(T-70)dt=0, T(0)=140`

We have to find a particular solution that satisfies the initial condition.

We can write,

`dT=-K(T-70)dt`

`\frac{dT}{T-70}=-Kdt`

`\int \frac{dT}{T-70}=\int -Kdt`

`ln(T-70)=-Kt+C` where C is a constant.

Now,

`T-70=e^{-Kt+C}`

`=e^{-Kt}.e^{C}`

`=C'e^{-Kt}` where `e^C=C'` is again a constant.

Hence we have,

`T=70+C'e^{-Kt}`

Applying the initial condition we get,

`140=70+C' ` implies `C'=70`

Therefore we have the solution:

`T=70(1+e^{-Kt})`