# Exercise 44

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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 |

Find the differential equations satisfied by the following families of curves.

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) | \[y^{2}=\frac{x^{3}}{c-x}\] |

Solve the equations:

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 [math]\frac{dL}{dt}=\frac{t}{72}\left(8-t\right)^{\frac{1}{3}}[/math] for [math]0 \leq t \leq 8[/math]. Here [math]L[/math] is the weight loss, as a percentage of the animal's original weight, [math]t[/math] weeks from the time of infection. Find the function giving the percentage weight loss [math]L[/math] at time [math]t[/math], 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 [math]W(t)[/math] be the weight of the cell at time [math]t[/math]. Assume that for a limited time the growth rate [math]\frac{dW}{dt}[/math] is proportional to [math]W^{\frac{2}{3}}[/math]. (If the density remains constant, then [math]W[/math] is proportional to [math]D^{3}[/math], where [math]D[/math] is the diameter of the cell, and the surface area is proportional to [math]D^{2}[/math], or equivalently [math]W^{\frac{2}{3}}[/math]. Hence, [math]\frac{dW}{dt}=kW^{\frac{2}{3}}[/math] for some positive constant [math]k[/math]. 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 [math]N[/math] is assumed to be given by the equation \[\frac{dW}{dt}=kW^{\frac{2}{3}}\], where [math]c[/math] and [math]K[/math] are positive constants. Find the formula giving the number of mice [math]t[/math] 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 [math]\frac{dq}{dt}=k\left(a-q\right)\left(b+q\right)[/math] where [math]q[/math] denotes the concentration of enzyme B produced after [math]t[/math] hours. Here [math]a[/math] and [math]b[/math] are the initial concentrations of enzyme A and B, respectively, while [math]k[/math] is a positive constant. If the initial concentrations, [math]a[/math] and [math]b[/math], respectively, are 100 and 5 moles/litre find the equation giving the concentration of enzyme B produced in the reaction after [math]t[/math] 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, [math]s[/math] in a population of total size [math]n+1[/math], become infected is given by [math]\frac{ds}{dt}=-cs\left(n+1-s\right)[/math] where [math]c[/math] 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:
\[\frac{cn\left(n+1\right)^{2}e^{\left(n+1\right)ct}}{\left(n+e^{\left(n+1\right)ct}\right)^{2}}\] |