Exercise 45

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Example (i) \[\frac{dy}{dx}+\frac{y}{x}=\frac{1}{x^{2}}\]
Example (ii) \[\frac{dy}{dx}+y\cot{x}=\sec{x}\]

Question 1(i) \[\frac{dy}{dx}+\frac{y}{x}=4x^{2}\]
Question 1(ii) \[2\left(y-4x^{2}\right)+x\frac{dy}{dx}=0\]
Question 1(iii) \[\frac{dy}{dx}+xy=x\]
Question 1(iv) \[\left(x^{5}+3y\right)-x\frac{dy}{dx}=0\]
Question 1(v) \[\frac{dy}{dx}+\frac{2y}{x^{2}-1}=\frac{x^{2}}{x-1}\]
Question 1(vi) \[\frac{dy}{dx}\cos{x}-y\sin{x}+\cot{x}=0\]
Question 1(vii) \[\frac{dy}{dx}+y\cot{x}=\sin{2x}\]
Question 1(viii) \[\frac{dy}{dx}=x^{3}-2xy\]
Question 1(ix) \[\frac{dy}{dx}=2\left(2x-y\right)\]
Question 1(x) \[\frac{dy}{dx}-my=ae^{2mx}\]
Question 1(xi) \[x\ln{x}\frac{dy}{dx}+y=2x^{2}\]
Question 1(xii) \[x\left(x^{2}+1\right)\frac{dy}{dx}+2y=\left(x^{2}+1\right)^{3}\]
Question 1(xiii) \[L\frac{di}{dt}+Ri=E\]
Question 1(xiv) \[\frac{dy}{dx}-my=ae^{mx}\]
Question 2 The population of a country changes at a rate proportional to the current size of the population and the prevailing economic situation. If the economy has a ten-year cycle, resulting in periods of immigration or emigration

\[\frac{dN}{dt}=kN+c\sin{\left(\frac{\pi t}{5}+a\right)}\] where [math]N[/math] is the size of the population at time [math]t[/math], measured in years, and [math]k[/math], [math]a[/math] and [math]c[/math] are constants.

Find an expression for the size of the population at any time.

Question 3 Verify that the differential equation for logistic growth [math]\frac{dN}{dt}=cN(K-N)[/math] can be converted into the linear differential equation [math]\frac{dx}{dt}+cKx=c[/math] by replacing [math]N[/math] by [math]\frac{1}{x}[/math].

\[x=\frac{1}{N}\] \[\frac{dx}{dt}=-\frac{1}{N^{2}}\frac{dN}{dt}\] Obtain the equation of logistic growth by first solving the differential equation for [math]x[/math] as a function of [math]t[/math] and then reintroducing [math]N[/math].

Question 4 A modification to the logistic equation which accounts for the influence of the resources available on the growth of the population results in the equation

\[\frac{dN}{dt}=c\frac{\left(K-N\right)N}{K+bN}\] where [math]b[/math], [math]c[/math] and [math]K[/math] are positive constants. Determine the solution of this equation, given that [math]N=N_0[/math] when [math]t=0[/math].

Show that the population size [math]N[/math] tends to the carrying capacity of the environment [math]K[/math] as [math]t[/math] increases.