Answer:
[tex]y(t)=25+Ce^{-\frac{t}{150}}[/tex]
Step-by-step explanation:
We are given that differential equation
[tex]\frac{dy}{dt}=-\frac{1}{50}(y-25)[/tex]
We have to find the expression for the temperature of the coffee at time t.
Let y be the temperature of the coffee in degree C and t be the time in minutes.
At t=0 , y=85 degree Celsius
[tex]\frac{dy}{y-25}=-\frac{1}{50}dt[/tex]
Integrating on both sides
[tex]\int \frac{dy}{y-25}=-\frac{1}{50}\int dt[/tex]
[tex]ln(y-25)=-\frac{t}{50}+lnC[/tex]
[tex]ln (y-25)-ln C=-\frac{t}{50}[/tex]
[tex]ln(m)-ln (n)=ln\frac{m}{n}[/tex]
[tex]ln\frac{y-25}{C}=-\frac{t}{50}[/tex]
[tex]\frac{y-25}{C}=e^{-\frac{t}{50}}[/tex]
[tex]lnx=y\implies x=e^y[/tex]
[tex]y-25=Ce^{-\frac{t}{150}}[/tex]
Substitute the value t=0 and y=85 then we get
[tex]85-25=Ce^{0}[/tex]
[tex]60=C[/tex]
Substitute the value of C
Then , we get
[tex]y-25=60e^{-\frac{t}{150}}[/tex]
[tex]y(t)=25+Ce^{-\frac{t}{150}}[/tex]
This is required expression for the temperature of the coffee at time t.
