Why does C3(t) = that?
OK, then read very carefully, for I will post this only once...
To understand the Integrating Factor method, you need to have a basic idea of the Product Rule, i.e. if $\displaystyle \displaystyle u,v$ are functions of $\displaystyle \displaystyle x$, then $\displaystyle \displaystyle \frac{d}{dx}[u\,v] = u\,\frac{dv}{dx} + v\,\frac{du}{dx}$.
Now examine a general first-order linear ODE:
$\displaystyle \displaystyle \frac{dy}{dx} + P(x)\,y = Q(x)$.
What we aim to do is to multiply both sides of the DE by some function of $\displaystyle \displaystyle x$, call it $\displaystyle \displaystyle I(x)$, so that the LHS is a Product-Rule expansion, which can then be written as a derivative, and therefore be integrated. BTW $\displaystyle \displaystyle I(x)$ is known as the Integrating Factor (hence the name of the method)
So multiplying both sides by $\displaystyle \displaystyle I(x)$ gives
$\displaystyle \displaystyle I(x)\,\frac{dy}{dx} + I(x)P(x)\,y = I(x)Q(x)$.
The LHS will be a Product Rule expansion if $\displaystyle \displaystyle I(x)P(x) = \frac{dI}{dx}$ (Can you see why?)
This is a separable ODE, so solving for $\displaystyle \displaystyle I$ we have
$\displaystyle \displaystyle P(x) = \frac{1}{I}\,\frac{dI}{dx}$
$\displaystyle \displaystyle \int{P(x)\,dx} = \int{\frac{1}{I}\,\frac{dI}{dx}\,dx}$
$\displaystyle \displaystyle \int{P(x)\,dx} = \int{\frac{1}{I}\,dI}$
$\displaystyle \displaystyle \int{P(x)\,dx} = \ln{I}$ (You can omit modulus signs and constants because any scalar multiple of the Integrating Factor will do)
$\displaystyle \displaystyle I = e^{\int{P(x)\,dx}}$.
So, if we multiply both sides of a first-order linear ODE by $\displaystyle \displaystyle e^{\int{P(x)\,dx}}$, we create a Product Rule expansion on the LHS which we can write as a single derivative and integrate both sides. Watch what happens...
$\displaystyle \displaystyle e^{\int{P(x)\,dx}}\,\frac{dy}{dx} + e^{\int{P(x)\,dx}}\,P(x)\,y = e^{\int{P(x)\,dx}}\,Q(x)$
$\displaystyle \displaystyle \frac{d}{dx}(e^{\int{P(x)\,dx}}\,y) = e^{\int{P(x)\,dx}}\,Q(x)$
$\displaystyle \displaystyle e^{\int{P(x)\,dx}}\,y = \int{e^{\int{P(x)\,dx}}\,Q(x)\,dx}$
$\displaystyle \displaystyle y = e^{-\int{P(x)\,dx}}\int{e^{\int{P(x)\,dx}}\,Q(x)\,dx}$
which solves the ODE.
Looks complicated in theory, in practice, extremely easy. Look at your problem...
$\displaystyle \displaystyle \frac{dC}{dt} + 9C = e^{-t}$.
Here your $\displaystyle \displaystyle P(x) = 9$, so the Integrating Factor is $\displaystyle \displaystyle e^{\int{9\,dt}} = e^{9t}$.
Multiplying both sides by the Integrating Factor gives
$\displaystyle \displaystyle e^{9t}\,\frac{dC}{dt} + 9e^{9t}C = e^{8t}$
$\displaystyle \displaystyle \frac{d}{dt}(e^{9t}C) = e^{8t}$
$\displaystyle \displaystyle e^{9t}C = \int{e^{8t}\,dt}$
$\displaystyle \displaystyle e^{9t}C = \frac{1}{8}e^{8t} + c$ where $\displaystyle \displaystyle c$ is the Integration Constant
$\displaystyle \displaystyle C = \frac{1}{8}e^{-t} + ce^{-9t}$.
It doesn't. You must have misunderstood the question. $\displaystyle Ae^{-t}$ is a solution to that equation only for $\displaystyle A= 1/8$. Since you have NOT taken a course in differential equations, I suspect you were NOT intended to actually solve that equation. I suspect that the problem was really "Find a value of A such that $\displaystyle C_3(t)= Ae^{-t}$ satisfies the equaton $\displaystyle C_3'(t)+ 9C_3(t)= e^{-t}$".
And you can do that just by differentiating the given function and plugging into the equation: if $\displaystyle C_3(t)= Ae^{-t}$ then $\displaystyle C_3'(t)= -Ae^{-t}$ so the equation becomes $\displaystyle -Ae^{-t}+ 9Ae^{-t}= e^{-t}$. That is the same as $\displaystyle 8Ae^{-t}= e^{-t}$ and since an exponential is never negative, we can divide both sides of the equation by $\displaystyle e^{-t}$ to get $\displaystyle 8A= 1$ or $\displaystyle A= \frac{1}{8}$.