From Mathematics Digital Library
Jump to navigation
Jump to search
| 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:
\[\frac{cn\left(n+1\right)^{2}e^{\left(n+1\right)ct}}{\left(n+e^{\left(n+1\right)ct}\right)^{2}}\]
|
