I have the following differential equation
$my'' + \beta y' + mg = 0$ , with condition $y(0)=0$.
I need to solve the equation and obtain expressions for $y(t)$ and $y'(t)$.
I have attempted to use the try method of substituting in $y=e^{\lambda t}$ but do not know how to proceed beyond that.
The differential equation is meant to represent free-fall of an object with air resistance that is proportional to it's speed.
The equation is a second-order linear, inhomogenous ordinary differential equation with constant coefficients. You can accept or search after it, but the general solution can be given in the form $y(t) = y_h (t) + y_p (t)$, where $y_h (t)$ is called the solution to the homogeneous differential equation (homogeneous means there is no constant term in the equation, in this example the $mg$ term), while $y_p(t)$ is called one particular solution to the original (inhomogeneous) equation.
Beginning with $y_h (t)$:
\begin{equation} my'' + \beta y' =0, \end{equation} This is the homogeneous differential equation. If you substitute $e^{\lambda t}$, you can get the value of $\lambda$ (which should not depend on t, of course). This comes from the fact that $\frac{d}{dt} e^t = e^t$. You'll get \begin{equation} \text{something} \cdot e^{\lambda t} = 0, \end{equation} and as it is valid for any t (and it is never 0), you can divide by $e^{\lambda t}$. You will get 2 possible $\lambda$-s, you will need both: $y_h (t) = C_1 \cdot e^{\lambda_1 t}+ C_2 \cdot e^{\lambda_2 t}$.
For $y_p(t)$, you only need one particular solution. If you see one immediately, in any form, you are lucky. I suggest looking for a solution of the form $y_p (t) = A + Bt$.
Once you got $y_h$ and $y_p$, you simply add the two together and get the general solution. Finally, use the initial conditions to determine the two constants.