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I was trying to do my physic homework and I got stuck here: $$m\ddot z + b\dot z + mg =0 $$ We know that $z(0) =z_0$ and $\dot z(0)=0$.

It is also suggested to use $\tau \equiv \frac m b$

I've added the teacher's answer below.

This is the answer the teacher has

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    We haven't seen them in math class yet, so if you could give me a hint on how to solve them I would be really grateful. – 2017-01-19
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    Have you seen first order ODEs? First divide through by $m$ and call $\frac{m}{b}=\tau$. Then use substitution the of $u=\dot{y}$ and $u'=\ddot{y}$. – 2017-01-19
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    Ok what do you do once you've got $ \tau \dot u + u +\tau g =0 $ ? @emka – 2017-01-19
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    u use a change of variable to get something like: u˙+ku= 0 the solution to that is exp(-kt + const) – 2017-01-19
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    By the way, does anyone know what to do once a question has been answered ? How do I close this thread ? – 2017-01-19
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    @HowCanIHelpYou There is no need to close the thread. Just accept the answer which helped you the most. We don't close threads here, because that way, if you find a better answer to the question you can accept that one instead. – 2017-01-19

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Feel free to point out any errors as the morning coffee hasn't kicked in.

Divide through by $m$ and use the substitution $\tau = \frac{b}{m}$:

$\ddot{z}+\tau \dot{z}+g=0$

Then use the substitution $v=\dot{z}$. Then $\dot{v}=\ddot{z}$

$\dot{v}+\tau v+g=0$
$\dot{v}+\tau v=-g$

Now that you have a first order ODE, just use an integrating factor: $\mu(t)=e^{\int \tau dt}=e^{\tau t}.$

$e^{\tau t}\dot{v}+e^{\tau t}\tau v=-g$

$(e^{\tau t}v)'=-e^{\tau t}g$

Integrate both sides w.r.t. $t$

$v=\frac{-e^{\tau t}g}{\tau}+c$

$\dot{z}=\frac{-e^{\tau t}g}{\tau}+c$

Integrating once more w.r.t $t$:

$z=\frac{-e^{\tau t}g}{\tau^2}+ct+d$

To find $c$ and $d$, plug in the conditions given.

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Let's solve the ODE,

$\displaystyle \ddot x + a\dot x+b=0$ where $a,b$ are arbitrary real constants. It's a homogeneous $2^{\text{nd}}$ order differential equation which has a corresponding A.E(Auxiliary Equation) given by,

$\displaystyle x^2+ax+b=0$ , Let it's roots be real and distinct (which is evident here as $b^2\ne 4m^2 g$) and denoted by $\alpha,\beta$.

So the solution of the ODE is given by $\displaystyle x(t)=c_1e^{\alpha t}+c_2e^{\beta t}$

In this case $\displaystyle a=\frac{b}{m},b=g$ and boundary conditions are $z(0)=z_0,\dot z(0)=0$ plugging in which we find ,

$\displaystyle z(t)=\frac{z_0}{\beta-\alpha}\left(\beta e^{\alpha t}-\alpha e^{\beta t}\right)$

where $\displaystyle \alpha,\beta = \frac{-b\pm\sqrt{b^2-4gm}}{2m}$

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    I think you've made a mistake using the auxillary equation, since it's actually non-homogeneous. Note that it isn't in the form $x''+ax'+b\color{red}{x}=0$ – 2017-01-19
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We found the answer. You just have to do as @emka said, then once you got: $$ \dot u +\frac {1}{\tau} u=0$$

you know that the general answer will be : $$ u(t) = C e^{(-\frac {t}{\tau} )}$$ You find $C = -\tau $ then you'll have: $$ \dot z(t) = \tau g (e^{(-\frac {t}{\tau})}-1)$$

Then all you have to do is integrate from 0 to t, and use you conditions to get the final equation.