I saw a problem similar to this in another question posted here. Unfortunately the question was put on hold because the poster didn't show what they had attempted. I thought it was a neat example of a differential equation so I changed the values and I'm asking it here.
A $30$ kg block is released from the top of a plane inclined at $30$ degrees to the horizontal. As the block slides down the plane, its coefficient of friction is $.30$, and it experiences a drag force in Newtons due to air resistance equal to $5$ times its velocity in m/s. Determine the velocity and displacement at any time t and after $5$ seconds after it is released.
A diagram is always helpful to determine the resultant forces and to formulate the differential equation. For a $30$ kg mass, the force due to its weight is $30\cdot 9.8 = 294$ N
$a = \frac{F}{m}$
$\frac{dV}{dt} = \frac{147 - (0.3\cdot 147\sqrt3 + 5V)}{30}$
$\frac{dV}{dt} = \frac{14.1233 - V}{6}$
$6\frac{dV}{dt} = 14.1233 - V$
$6\frac{dV}{14.1233 - V} = dt$
$-6\frac{dV}{V - 14.1233} = dt$
$-6\ln|V - 14.1233| = t + C$
$\ln|V - 14.1233| = \frac{-t-C}{6}$
$V - 14.1233 = e^{-\frac{C}{6}}\cdot e^{-\frac{t}{6}}$
$V = 14.1233 + e^{-\frac{C}{6}}\cdot e^{-\frac{t}{6}}$
When $t = 0, V = 0$
$0 = 14.1233 + e^{-\frac{C}{6}}$
$e^{-\frac{C}{6}} = -14.1233$
$V = 14.1233 - 14.1233\cdot e^{-\frac{t}{6}}$ is the solution
When $t = 5$
$V = 14.1233 - 14.1233\cdot e^{-\frac{5}{6}}$
$$V = 7.9853\text{ m/s}$$
$\frac{dS}{dt} = 14.1233 - 14.1233\cdot e^{-\frac{t}{6}}$
$S = 84.7398e^{-\frac{t}{6}} + 14.1233t + C$
When $t=0, S = 0$
$C = -84.7398$
$S = 84.7398e^{-\frac{t}{6}} + 14.1233t -84.7398$ is the solution
When $t = 5$
$S = 84.7398e^{-\frac{5}{6}} + 14.1233(5) -84.7398$
$$S = 22.7045\text{ m}$$