Question

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

Answer 1

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

Note:
We can solve the question by separating the variable. First we divide LHS and RHS by T-70, after dividing we get $\dfrac{dT}{dT-70}+kdt=0$ . We know that integration of 0 is a constant. Now integrating both sides we get $\ln \left( T-70 \right)+kt=c$ 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 $T=70+2{{e}^{-kt}}$