General solution of $T'(t) = \mu T(t)$?

Question

So I'm trying to find the general solution that satisfy a given partial differential equation and boundary conditions via the method of separation of variables.

However, I cannot work out or find anywhere an explanation of how to find the general solution of something in the format $T'(t) = \mu T(t)$? Perhaps I'm missing something obvious or forgetting a topic I've learnt previously, but I can't work it out.

I am aware how to find the general solution of something in the form $T''(t) = \mu T(t)$, which depends on the value of $\mu$. i.e. if $\mu$ is negative, then the general solution takes the form $T(t) = Asin(kx)+Bcos(kx)$. I just can't work out what to do when the equation only involves a first-order differential?

I'm not looking for answers to my particular problem, hence why I've not provided any specific values - just an explanation of the form the general solution takes when the equation is a first-order differential. Thanks in advance!

Answer 1

The solution to this equation is well known and easy to verify:

$T(t)=Ce^{\mu t}$

Answer 2

Your differential equation in Leibniz notation is: $$\frac{dT}{dt}=\mu T$$ This is a separable differential equation, since it is in the form: $$\frac{dy}{dx}=f(x)g(y)$$ In general, these can be solved by dividing both sides by $g(y)$ and then integrating both sides with respect to $x$: $$\int \frac{1}{g(y)}~dy=\int f(x)~dx$$

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Therefore, in your specific case, you can divide both sides of the equation by $T$, and then integrate both sides with respect to $t$. This gives: $$\int \frac{1}{T}~dT=\int \mu ~dt$$ Integrating both sides, we obtain: $$\ln|T|=\mu t+C$$ Exponentiating both sides, we obtain the general solution. Note that the absolute value sign can be omitted. $$T=e^{\mu t+C}$$ $$T=ke^{\mu t}$$ Where $k=e^C$.

Answer 3

Divide by $T(t)$: $$\frac{T'(t)}{T(t)}=\mu$$Integrate both sides with respect to $t$:$$\ln|T(t)|=\mu t+C$$(Using $u$-substitution). Raise both sides to $e$: $$T(t)=e^{\mu t +C}=e^Ce^{\mu t}=ce^{\mu t}$$This answer is like @projectilemotion's but the first step is different.