# Exercise 44

 Example (i) $\frac{dy}{dx}=\frac{x}{y}$ Solution Example (ii) $\frac{dy}{dx}=\frac{y\left(1+x\cot{x}\right)}{x}$ Solution Example (iii) $\frac{dy}{dx}=\frac{y^{2}}{1+x^{2}}$ Solution
 Question 1(i) $x^{3}-3x^{2}y=c$ Question 1(ii) $x^{2}y=1+cx$ Question 1(iii) $y=6+ce^{2x}$ Question 1(iv) $cy^{2}=x^{2}\left(x-2\right)$ Question 1(v) $y^{2}=\frac{x^{3}}{c-x}$
 Question 2(i) $\frac{dy}{dx}+x=\sin{x}$ Question 2(ii) $\frac{dy}{dx}=y\left(y-1\right)$ Question 2(iii) $x\left(x+1\right)\frac{dy}{dx}=2$ Question 2(iv) $x^{2}y\frac{dy}{dx}=1$ Question 2(v) $\left(x+1\right)\frac{dy}{dx}=y+3$ Question 2(vi) $x\left(x+2\right)\frac{dy}{dx}=e^{y}$ Question 2(vii) $xy\frac{dy}{dx}=2$ Question 2(viii) $x^{2}e^{y}\frac{dy}{dx}=1$ Question 2(ix) $x\frac{dy}{dx}+y=xy$ Question 2(x) $\frac{dy}{dx}\cos{x}=y\sin{x}$ Question 2(xi) $y\frac{dy}{dx}-x\left(1-y^{2}\right)=0$ Question 2(xii) $\frac{dy}{dx}=3x^{2}y+3x^{2}$ Question 2(xiii) $x\frac{dy}{dx}=2y-xy\tan{x}$ Question 2(xiv) $\frac{dy}{dx}\sqrt{1-x^{2}}=x$ Question 2(xv) $\frac{dy}{dx}=xy^{3}$ Question 2(xvi) $\frac{dy}{dx}\left(1+x^{4}\right)=2xy^{2}$
 Question 3 The rate at which weight is lost by an animal suffering from a viral infection is observed to satisfy the relationship $\frac{dL}{dt}=\frac{t}{72}\left(8-t\right)^{\frac{1}{3}}$ for $0 \leq t \leq 8$. Here $L$ is the weight loss, as a percentage of the animal's original weight, $t$ weeks from the time of infection. Find the function giving the percentage weight loss $L$ at time $t$, and determine the percentage weight loss when the animal dies eight weeks later. Question 4 The growth of a cell depends on the flow of nutrients through its surface. Let $W(t)$ be the weight of the cell at time $t$. Assume that for a limited time the growth rate $\frac{dW}{dt}$ is proportional to $W^{\frac{2}{3}}$. (If the density remains constant, then $W$ is proportional to $D^{3}$, where $D$ is the diameter of the cell, and the surface area is proportional to $D^{2}$, or equivalently $W^{\frac{2}{3}}$. Hence, $\frac{dW}{dt}=kW^{\frac{2}{3}}$ for some positive constant $k$. Find the general solution to this differential equation. Question 5 A population of mice, initially numbering 30, is kept in conditions that can support a population size of 120. The rate of increase in the population size $N$ is assumed to be given by the equation $\frac{dW}{dt}=kW^{\frac{2}{3}}$, where $c$ and $K$ are positive constants. Find the formula giving the number of mice $t$ months later, if, after one month, the population numbers 80. Question 6 The rate at which enzyme A is converted to enzyme B is governed by the differential equation $\frac{dq}{dt}=k\left(a-q\right)\left(b+q\right)$ where $q$ denotes the concentration of enzyme B produced after $t$ hours. Here $a$ and $b$ are the initial concentrations of enzyme A and B, respectively, while $k$ is a positive constant. If the initial concentrations, $a$ and $b$, respectively, are 100 and 5 moles/litre find the equation giving the concentration of enzyme B produced in the reaction after $t$ hours when the concentration of enzyme B produced in the first hour is 10 moles/litre. Question 7 In a 'simple' epidemic the rate at which the number of susceptibles, $s$ in a population of total size $n+1$, become infected is given by $\frac{ds}{dt}=-cs\left(n+1-s\right)$ where $c$ is a positive constant. Show that, if an epidemic is 'simple', the number of new cases of the disease recorded per unit time will be give by: $\frac{cn\left(n+1\right)^{2}e^{\left(n+1\right)ct}}{\left(n+e^{\left(n+1\right)ct}\right)^{2}}$