# First order seperable ode with IC's

• wtmoore
In summary, according to Newton’s Law of Cooling, the rate at which a substance cools in air is proportional to the difference between the temperature of the substance and that of air, as shown by the differential equation dT/dt=-k(T-T​0). The general solution for this equation is T=Cexp(-kt)+T0. When T0=30°C, the particular solution is T=70exp(-kt)+30, with T=100°C and t=0 as the initial condition. To determine the physical constant k, an experiment is carried out where the substance cools from 100°C to 70°C in 15 minutes, resulting in k=0.0373. The time taken for the temperature
wtmoore

## Homework Statement

Question

According to Newton’s Law of Cooling, the rate at which a substance cools in air is proportional to the difference between the temperature of the substance and that of air. The differential equation is given byAccording to Newton’s Law of Cooling, the rate at which a substance cools in air is proportional to the difference between the temperature of the substance and that of air. The differential equation is given by:

dT/dt=-k(T-T​0)

where T is the temperature of the substance, T0 is the temperature of air, k is a constant of proportionality and t is the time. Find the general solution of this equation.
If T0=30°C then find the solution subject to the initial condition that T=100°C at time t=0.
Suppose an experiment is carried to determine the physical constant k. If the substance cools from 100°C to 70°C in 15 minutes then find k. Hence find the time taken for the temperature to fall from 100°C to 40°C

## The Attempt at a Solution

Ok guys, Just so you know where I'm at, I can do seperable ODES quite easily, however, I am useless and wordy questions and I don't know how to set it up with IC's in. I need to bring the T under the dT, but don't know how to separate it from -k(T-T0).

If I multiply out I get

dT/dt=-kT+T0

and I know know how to get it under the dT, can someone explain how I do this? I may need further help with the initial conditions but I need to know how to find the general solution for this type of ode with IC's first.

Writing it in the form dT/(T-T0)=(-k)dt is separated enough. T0 and k are just constants. You can integrate both sides. Use a trivial change of variables on the dT/(T-T0) side.

Ok,

so now the general solution is:

T=Cexp(-kt)+T0

particular solution is

T=70exp(-kt)+30 when T=100 when t=0 and T0=30

However for the next bit I don't understand,

Am I doing T(15)=70?

Does this give

T=Cexp(-15k)+30?

So 70=70exp(-15k)+30

k=0.0373

Then T=40

40=70exp(-0.0373t)+30

t=52mins?

Last edited:
wtmoore said:
Ok,

so now the general solution is:

T=Cexp(-kt)+T0

particular solution is

T=70exp(-kt)+30 when T=100 when t=0 and T0=30

However for the next bit I don't understand,

Am I doing T(15)=70?

Does this give

T=Cexp(-15k)+30?

So 70=70exp(-15k)+30

k=0.0373

Then T=40

40=70exp(-0.0373t)+30

t=52mins?

Looks ok to me.

thanks dick

## 1. What is a first order separable ODE?

A first order separable ODE (ordinary differential equation) is a type of differential equation where the dependent variable and its derivative can be separated on opposite sides of the equation, allowing it to be solved through integration.

## 2. What are "IC's" in relation to first order separable ODE?

IC's stands for initial conditions and refers to the values of the dependent variable and its derivative at a specific point in the domain of the equation. These values are used to find the particular solution of the ODE.

## 3. How do you solve a first order separable ODE with IC's?

To solve a first order separable ODE with IC's, you need to first separate the variables and then integrate both sides of the equation. Then, substitute in the initial conditions to find the particular solution.

## 4. What is the general solution of a first order separable ODE?

The general solution of a first order separable ODE is the collection of all possible solutions that satisfy the equation. It is usually expressed in terms of a constant, which can be determined by substituting in the initial conditions.

## 5. What are some real-life applications of first order separable ODEs?

First order separable ODEs are commonly used in physics, biology, and economics to model various real-life phenomena. Examples include population growth, radioactive decay, and the spread of infectious diseases.

Replies
4
Views
951
Replies
6
Views
2K
Replies
3
Views
1K
Replies
16
Views
1K
Replies
2
Views
1K
Replies
5
Views
1K
Replies
14
Views
3K
Replies
25
Views
2K
Replies
2
Views
2K
Replies
2
Views
1K