# dT + k(T-70)dt = 0 , T(0) = 140 Find the particular solution that satisfies the initial condition

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})

Approved by eNotes Editorial Team