What is a solution to the differential equation #dT+k(T-70)dt# with T=140 when t=0? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Eddie Jul 18, 2016 #T(t) = 70 (1 + e^(-kt) )# Explanation: #dT+k(T-70)dt color{red}{= 0}# Ahem, that was missing this is separable #(dT)/(dt)= - k(T-70)# #1/(T-70) (dT)/(dt)= - k# #int \ 1/(T-70) (dT)/(dt) \ dt = - k int \ dt# #int \ 1/(T-70) dT = - k int \ dt# #ln (T-70) = - kt + C# #T-70 = e^ (- kt + C) = C e^(-kt)# Applying IV #140-70 = C e^(-k(0)) implies C = 70# #T(t) = 70 (1 + e^(-kt) )# Answer link Related questions How do you solve separable differential equations? How do you solve separable first-order differential equations? How do you solve separable differential equations with initial conditions? What are separable differential equations? How do you solve the differential equation #dy/dx=6y^2x#, where #y(1)=1/25# ? How do you solve the differential equation #y'=e^(-y)(2x-4)#, where #y5)=0# ? How do you solve the differential equation #(dy)/dx=e^(y-x)sec(y)(1+x^2)#, where #y(0)=0# ? How do I solve the equation #dy/dt = 2y - 10#? Given the general solution to #t^2y'' - 4ty' + 4y = 0# is #y= c_1t + c_2t^4#, how do I solve the... How do I solve the differential equation #xy'-y=3xy, y_1=0#? See all questions in Solving Separable Differential Equations Impact of this question 3086 views around the world You can reuse this answer Creative Commons License