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Question: How do you find the particular solution to \(dT+k\left( T-70 \right)dt=0\) that satisfies \(T\left( ...

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

Explanation

Solution

We can see that the above equation is a variable separable equation that means we can separate T and t and integrate them in LHS and RHS. Integration of dyy−c\dfrac{dy}{y-c} equal to ln⁡(y−c)\ln \left( y-c \right) where c is a constant. We will use this to solve the question.

Complete step by step answer:
The differential equation given in the question dT+k(T−70)dt=0dT+k\left( T-70 \right)dt=0 and T(0)=140T\left( 0 \right)=140
We can see that we can separate T and t
dTT−70=−kdt\dfrac{dT}{T-70}=-kdt
Now integrating both sides we get
∫dTT−70=−k∫dt\int{\dfrac{dT}{T-70}}=-k\int{dt}
We know that integration of dyy−k=ln⁡(y−k)\dfrac{dy}{y-k}=\ln \left( y-k \right) and integration of 1dt is t . It is a indefinite integration so we have to add a constant in the end of solution
ln⁡(T−70)=−kt+c\ln \left( T-70 \right)=-kt+c
Now we have to calculate the constant using T(0)=140T\left( 0 \right)=140 if we put t=0 then we will get the value of T as 140
So ln⁡(140−70)=0+c\ln \left( 140-70 \right)=0+c
So the value of c is equals to ln2
So we can write ln⁡(T−70)=−kt+ln⁡2\ln \left( T-70 \right)=-kt+\ln 2
Taking power of e in LHS and RHS we get
T−70=2e−ktT-70=2{{e}^{-kt}}
Adding 70 in LHS and RHS of the equation
T=70+2e−ktT=70+2{{e}^{-kt}}
So the particular solution for the equation is T=70+2e−ktT=70+2{{e}^{-kt}}

Note:
We can solve the question by separating the variable. First we divide LHS and RHS by T-70, after dividing we get dTdT−70+kdt=0\dfrac{dT}{dT-70}+kdt=0 . We know that integration of 0 is a constant. Now integrating both sides we get ln⁡(T−70)+kt=c\ln \left( T-70 \right)+kt=c where c is a constant. We can find the value of c by putting T=140 and t=0, we will get the value of c as ln2. Then we can take power of e in both LHS and RHS and we will get T=70+2e−ktT=70+2{{e}^{-kt}}