A chemical substance $A$ changes into substance $B$ at rate $\alpha$ times the amount of $A$ present. Substance $B$ changes into $C$ at rate $\beta$ times the amount of $B$ present. If, initially, only substance $A$ is present and its amount is $M$, show that the amount of $C$ present at time $t$ is
$$M+M\left ( \frac{\beta e^{-\alpha t}-\alpha e^{-\beta t}}{\alpha - \beta}\right ) \tag{ 1}$$
So $a_0=M, b_0=0$ and $c_0=0$. Now I was doing some research, and found out that in these kinds of computations, the rate which amount of substance is decreasing over time are these solvable (by me) differential equations
$$\frac{dA}{dt}=-\alpha A \implies a_0e^{-\alpha t} = Me^{-\alpha t}\tag{2 }$$ $$\frac{dB}{dt}=-\beta B \implies b_0e^{-\beta t}\tag{3 }$$
The rate which the substances change formulas are these (they have something to do with the movement of particle in viscosious fluid)
$$\frac{dB}{dt} = \alpha A \implies b_0+a_0(1-e^{-\alpha t})\tag{ 4}$$ $$\frac{dC}{dt} = \beta B \implies b_0+a_0(1-e^{-\beta t})\tag{5 }$$
But I do not know how to move further to the final formula. Thanks for help in advance!
EDIT B from the answer:
$$ke^{-\beta t}-\frac{M\alpha e^{\beta t - \alpha t} e^{-\beta t}}{\alpha - \beta}\tag{6 }$$
Do not know what to do with $k$ now, and I cannot substitute a like in previous step.
If I put $k = b_0 = 0$ and simplify then, it becomes
$$-\frac{M\beta e^{\alpha t}}{\alpha - \beta}$$
Doing $C=M-(A+B)$ will result in
$$M-M e^{-\alpha t}+\frac{M \alpha e^{-\alpha t}}{\alpha - \beta}$$
I am still missing something.
Hint
Since you have $A\to B\to C$, I suppose that the kinetic model is $$\frac{dA}{dt}=-\alpha A$$ $$\frac{dB}{dt}=\alpha A-\beta B$$ $$\frac{dC}{dt}=\beta B$$ (remember that $B$ appears and disappears).
Integrating the first equation is simple. Using the initial condition, you obtained $$A=M e^{-\alpha t}$$ which is perfectly correct. So, the second equation write $$\frac{dB}{dt}=\alpha M e^{-\alpha t}-\beta B$$ which is not very complex and I let it to you.
For $C$, the problem is simple if you notice that $$\frac{dA}{dt}+\frac{dB}{dt}+\frac{dC}{dt}=0$$ which implies that, at any time $t$, $A+B+C=M$.
I am sure that you can take from here.