# Linear/Separable Differential Equations Question (Engineering First Year Calculus)?

## I have no idea how to solve this question. If someone could do it or provide links to their answer, that would be great thanks :)

May 24, 2018

Starting with the first bit

#### Explanation:

Part (a)

Velocity

$\frac{\mathrm{dv}}{\mathrm{dt}} = - k {v}^{n}$, with $k > 0$, and $n$ as some constant

Separate:

$\frac{\mathrm{dv}}{{v}^{n}} = - k \setminus \mathrm{dt}$

Excluding $n = 1$, that solves as:

$\frac{{v}^{1 - n}}{1 - n} = - k t + C$

With IV : $q \quad v \left(0\right) = {v}_{o}$

$\frac{{v}_{o}^{1 - n}}{1 - n} = C$

$\frac{{v}^{1 - n}}{1 - n} = - k t + \frac{{v}_{o}^{1 - n}}{1 - n}$

${v}^{1 - n} = {v}_{o}^{1 - n} - k \left(1 - n\right) t$

If $v \left(\tau\right) = 0$:

$\tau = \frac{1}{k} \cdot \frac{{v}_{o}^{1 - n}}{1 - n}$

We expect that $\tau > 0$, or the model is meaningless. So:

$\frac{1}{k} \cdot \frac{{v}_{o}^{1 - n}}{1 - n} > 0$

The givens:

• $k > 0 \implies \frac{1}{k} > 0$

• ${v}_{0} > 0 \implies {v}_{o}^{1 - n} > 0$

$\implies n < 1$

Displacement

$\frac{\mathrm{dv}}{\mathrm{dt}} = v \frac{\mathrm{dv}}{\mathrm{dx}} = - k {v}^{n} , \quad n < 1$

${v}^{1 - n} \mathrm{dv} = - k \mathrm{dx}$

$\frac{{v}^{2 - n}}{2 - n} = - k x + C , q \quad \left[n \ne 2\right]$

With IV:

• $x \left({v}_{o}\right) = 0$

$\implies C = \frac{{v}_{o}^{2 - n}}{2 - n}$

$x = \frac{1}{k \left(2 - n\right)} \left({v}_{o}^{2 - n} - {v}^{2 - n}\right)$

$v = 0 \implies x = \frac{{v}_{o}^{2 - n}}{k \left(2 - n\right)}$

The maximum distance depends upon the value of ${v}_{0}$. Eg representative values:

$x = \left\{\begin{matrix}n = - 1 & q \quad {v}_{0}^{3} / \left(3 k\right) \\ n = 0 & q \quad {v}_{0}^{2} / \left(2 k\right) \\ n = 1 & q \quad {v}_{0} / k\end{matrix}\right.$

[ Part (b), (c) - The rest of the question is answered in the above in one way or another]