# Laplace transform and inverse in Newton's law of cooling

#### codproplayd

##### New member
Hello,
I know that to integrate Newton's law of cooling's differential equation T′(t)=k(T(t)−A) where T=temperature, t=time, A=constant temperature of the surrounding, and k is constant, I can use Laplace transform:

T′(t)=K⋅(T(t)−A)→s⋅T(s)−T(0)=K⋅(T(s)−A⋅1/s)

Then solve for T (s) and use the inverse transform.

T(s)=(T(0)−K⋅A⋅(1/s))/(s−K) => T(t)=A+e^Kt(T(0)−A) correct me if I am mistaken please

And from there I could solve the equation, which is easy, but I don't know the step-by-step method of using both the Laplace transform and inverse with adding the integrals to clearly explain what I did on paper.
I would be thankful if someone would help me out with this one

#### HallsofIvy

##### Elite Member
Hello,
I know that to integrate Newton's law of cooling's differential equation T′(t)=k(T(t)−A) where T=temperature, t=time, A=constant temperature of the surrounding, and k is constant, I can use Laplace transform:

T′(t)=K⋅(T(t)−A)→s⋅T(s)−T(0)=K⋅(T(s)−A⋅1/s)

Then solve for T (s) and use the inverse transform.

T(s)=(T(0)−K⋅A⋅(1/s))/(s−K) => T(t)=A+e^Kt(T(0)−A) correct me if I am mistaken please
It looks like you have a sign wrong. From sT(s)- T(0)= K(T(s)- A/s)= KT(s)- KA/s, add T(0) to both sides and subtract KT(s): sT(x)- KT(s)= (s- K)T(s)= T(0)- KA/s so T(s)= (T(0)- KA/s)/(s- K). That is, "K- s" in the denominator, not "s- K".

And from there I could solve the equation, which is easy, but I don't know the step-by-step method of using both the Laplace transform and inverse with adding the integrals to clearly explain what I did on paper.
I would be thankful if someone would help me out with this one
I'm not clear what you mean by "the step-by-step method of using both the Laplace transform and inverse with adding the integrals". The Laplace transform of f(x) is $$\displaystyle F(s)= \int e^{-sx}f(x)dx$$. An integral is **linear**: $$\displaystyle \int af(x)+ bg(x) dx= a\int f(x)dx+ b\int g(x)dx$$. So the Laplace transform is also linear: $$\displaystyle L(af(x)+ bg(x))= \int_0^\infty e^{-sx}(af(x)+ bg(x) dx= a\int_0^\infty e^{-sx}f(x)dx+ b\int_0^\infty e^{-sx}g(x)dx= aL(f(x))+ bL(g(x))$$. Is that what you mean?

#### codproplayd

##### New member
Oh, I see.
Yes, this is what I exactly wanted. I just had to explain some steps in my maths report and lost train of thought.
Thank you so much