Answer:
[tex] T(t) = M -C_1 e^{-kt}[/tex]
And as we can see that represent the solution for the differential equation on this case.
Step-by-step explanation:
For this case we have the following differential equation:
[tex] \frac{dT}{dt}= k(M-T)[/tex]
We can rewrite this expression like this:
[tex] \frac{dT}{M-T} = k dt[/tex]
We can us the following susbtitution for the left part [tex] u = M-T[/tex] then [tex] du= -dt[/tex] and if we replace this we got:
[tex] \frac{-du}{u} = kdt[/tex]
We can multiply both sides by -1 and we got;
[tex] \frac{du}{u} =-k dt[/tex]
Now we can integrate both sides and we got:
[tex] ln |u| = -kt + C[/tex]
Where C is a contant. Now we can exponetiate both sides and we got:
[tex] u(t) =e^{-kt} *e^C = C_1 e^{-kt}[/tex]
Where [tex] C_1 = e^C[/tex] is a constant. And now we can replace u and we got this:
[tex] M-T = C_1 e^{-kt}[/tex]
And if we solve for T we got:
[tex] T(t) = M -C_1 e^{-kt}[/tex]
And as we can see that represent the solution for the differential equation on this case.
