Viscosity problem Suppose that the particle is subjected to a viscous resistive drag force proportional to the mass of the particle and its velocity: $F_D$=m$\alpha$$\frac{dr(t)}{dt}$ show that the equation becomes $\frac{d^2r}{dt^2}$=-$\alpha$$\frac{dr(t)}{dt}$-$k^2$r(t). Having some trouble where to start, any help will be appreciated

As far as i can see $\alpha$ and $k^2$ are constants.

We can rewrite the equation as:

$$r''(t)+\alpha r'(t)+k^2r(t)=0$$

This is a linear differential equation with constant coefficients. These types of differential equations can always be solved with the following Ansatz $r(t)=e^{\lambda t}$. Plug this into the equation to get the following:

$$(\lambda^2+\alpha\lambda+k^2)e^{\lambda t}=0$$

As $e^{\lambda t}>0$ the polynomial has to be equal to zero.

$\lambda^2+\alpha\lambda+k^2=0$

Solve this quadratic equation to get $\lambda_{1/2}$. The general solution can be expressed as

$$r(t)=c_1e^{\lambda_1 t}+c_2e^{\lambda_2 t}.$$