Problems

It is given that \(y\) satisfies $$ {{\d y} \over { \d t}} + k\left({{t^2-3t+2} \over {t+1}}\right)y = 0\;, $$ where \(k\) is a constant, and \(y=A \) when \(t=0\,\), where \(A\) is a positive constant. Find \(y\) in terms of \(t\,\), \(k\) and \(A\,\). Show that \(y\) has two stationary values whose ratio is \((3/2)^{6k}\e^{-5{k}/2}.\) Describe the behaviour of \(y\) as \(t \to +\infty\) for the case where \(k> 0\) and for the case where \(k<0\,.\) In separate diagrams, sketch the graph of \(y\) for \(t>0\) for each of these cases.

Show Solution

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Solution: \begin{align*} && \frac{\d y}{\d t} &= - k \left (\frac{t^2-3t+2}{t+1} \right) y \\ \Rightarrow && \int \frac1y \d y &= -k\int \left (t-4 + \frac{6}{t+1}\right) \d t \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + C \\ (t,y) = (0,A): && \ln A &=C \\ \Rightarrow && \ln y &= -k \left ( \frac12 t^2 -4t + 6\ln (t+1) \right) + \ln A \\ && \ln \left ( \frac{y}{A}(t+1)^{6k} \right) &= -k \l \frac12 t^2 - 4t \r \\ \Rightarrow && y &= A\frac{\exp \l -k(\frac12 t^2-4t)\r}{(t+1)^{6k}} \end{align*} \(y\) wil have stationary values when \(\frac{\d y}{\d t} = 0\), ie \begin{align*} k \left (\frac{t^2-3t+2}{t+1} \right) y &= 0 \\ k \left ( \frac{(t-2)(t-1)}{t+1} \right) y &= 0 \end{align*} ie when \(y = 0, t = 1, t =2\). Clearly \(y = 0\) is not a solution, so \(y\) has the values: \begin{align*} t = 1: && y &= A\frac{\exp \l -k(\frac12 -4)\r}{(2)^{6k}} \\ &&&= A \frac{e^{7/2 k}}{2^{6k}} \\ t = 2: && y &= A\frac{\exp \l -k(2 -8)\r}{(3)^{6k}} \\ &&&= A \frac{e^{6 k}}{3^{6k}} \\ \text{ratio}: && \frac{e^{7/2k}}{2^{6k}} \cdot \frac{3^{6k}}{e^{6k}} &= (3/2)^{6k} e^{-5k/2} \end{align*} If \(k > 0\) as \(t \to \infty\) \(y \to 0\) since the \(e^{-kt^2/2}\) term dominates everything. If \(k < 0\) as \(t \to \infty\) \(y \to \infty\) as since the \(e^{-kt^2}\) term also dominates but now it growing to infinity faster than everything else.

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Find \(y\) in terms of \(x\), given that: \begin{eqnarray*} \mbox{for \(x < 0\,\)}, && \frac{\d y}{\d x} = -y \mbox{ \ \ and \ \ } y = a \mbox{ when } x = -1\;; \\ \mbox{for \(x > 0\,\)}, && \frac{\d y}{\d x} = y \mbox{ \ \ \ \ and \ \ } y = b \ \mbox{ when } x = 1\;. \end{eqnarray*} Sketch a solution curve. Determine the condition on \(a\) and \(b\) for the solution curve to be continuous (that is, for there to be no `jump' in the value of \(y\)) at \(x = 0\). Solve the differential equation \[ \frac{\d y}{\d x} = \left\vert \e^x-1\right\vert y \] given that \(y=\e^{\e}\) when \(x=1\) and that \(y\) is continuous at \(x=0\,\). Write down the following limits: \ \[ \text{(i)} \ \ \lim_ {x \to +\infty} y\exp(-\e^x)\;; \ \ \ \ \ \ \ \ \ \text{(ii)} \ \ \lim_{x \to -\infty}y \e^{-x}\,. \]

Find all the solution curves of the differential equation \[ y^4 \l {\mathrm{d}y \over \mathrm{d}x }\r^{\! \! 4} = \l y^2 - 1 \r^2 \] that pass through either of the points

1. \(\l 0, \, \frac{1}{2}\sqrt3 \r\),
2. \(\l 0, \, \frac{1}{2}\sqrt5 \r\).

Show Solution

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Solution: \begin{align*} && y^4 \left (\frac{\d y}{\d x} \right)^4 &= (y^2 - 1)^2 \\ \Rightarrow && y^2 \left (\frac{\d y}{\d x} \right)^2 &= |y^2 - 1| \\ && y \left (\frac{\d y}{\d x} \right) &= \pm \sqrt{|y^2-1|} \\ \Rightarrow &&\int \frac{y}{\sqrt{|y^2-1|}} \d y &= \int \pm 1 \d x \\ \Rightarrow && \pm \sqrt{|y^2-1|} &= \pm x + C \\ \end{align*}

1. Since \(y^2 < 1\), our solution curve should be of the from \(-\sqrt{1-y^2} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{3})\), we obtain \(-\tfrac12 = C\), therefore our solution curves are \(\pm x = \frac12 - \sqrt{1-y^2}\)
2. Since \(y^2 > 1\), our solution curve should be of the from \(\sqrt{y^2-1} = \pm x + C\) Plugging in \((0, \tfrac12 \sqrt{5})\), we obtain \(\tfrac12 = C\), therefore our solution curves are \(\pm x = \sqrt{y^2-1}-\frac12\)

Clearly if \(y = \pm 1\), \(y'=0\) and the equation is satisfied.

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In a cosmological model, the radius \(\rm R\) of the universe is a function of the age \(t\) of the universe. The function \(\rm R\) satisfies the three conditions: $$ \mbox{\({\rm R}(0)=0\)}, \ \ \ \ \ \ \ \ \ \mbox{\({\rm R'}(t)>0\) for \(t>0\)}, \ \ \ \ \ \ \ \ \ \ \mbox{\({\rm R''}(t)<0\) for \(t>0\)}, \tag{*} $$ where \({\rm R''}\) denotes the second derivative of \(\rm R\). The function \({\rm H}\) is defined by \[ {\rm H} (t)= \frac{{\rm R}'(t)}{{\rm R}( t)}\;. \]

1. Sketch a graph of \({\rm R} (t)\). By considering a tangent to the graph, show that \(t<1/{\rm H}(t)\).
2. Observations reveal that \({\rm H}(t) = a/t\), where \(a\) is constant. Derive an expression for \({\rm R}(t)\). What range of values of \(a\) is consistent with the three conditions \((*)\)?
3. Suppose, instead, that observations reveal that \({\rm H}(t)= b t^{-2}\), where \(b\) is constant. Show that this is not consistent with conditions \((*)\) for any value of \(b\).

The function \(\f\) satisfies \(\f(x+1)= \f(x)\) and \(\f(x)>0\) for all \(x\).

1. Give an example of such a function.
2. The function \(\F\) satisfies \[ \frac{\d \F}{\d x} =\f(x) \] and \(\F(0)=0\). Show that \(\F(n) = n\F(1)\), for any positive integer \(n\).
3. Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} +\f(x) y=0 \] that satisfies \(y=1\) when \(x=0\). Show that \(y(n) \to 0\) as \(n\to\infty\), where \(n= 1,\,2,\, 3,\, \ldots\)

1. Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} + 4x\e^{-x^2} {(y+3)}^{\frac12} = 0 \qquad (x \ge 0), \] that satisfies the condition \(y=6\) when \(x=0\). Find \(y\) in terms of \(x\) and show that \(y\to1\) as \(x \to \infty\).
2. Let \(y\) be any solution of the differential equation \[ \frac{\d y}{\d x} -x\e^{6 x^2} (y+3)^{1-k} = 0 \qquad (x \ge 0). \] %that satisfies the condition \(y=6\) %when \(x=0\). Find a value of \(k\) such that, as \(x \to \infty\), \(\e^{-3x^2}y\) tends to a finite non-zero limit, which you should determine.

1. At time \(t=0\) a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time \(t\) is \(y\). Show that there is a constant \(b < 1\) such that \(y=b^{t}.\)
2. Suppose instead that the tank contains one unit of water at time \(t=0,\) but that in addition to water flowing out as described, water is added at a steady rate \(a>0.\) Show that \[ \frac{\mathrm{d}y}{\mathrm{d}t}-y\ln b=a, \] and hence find \(y\) in terms of \(a,b\) and \(t\).

Whenever I go cycling I start with my bike in good working order. However if all is well at time \(t\), the probability that I get a puncture in the small interval \((t,t+\delta t)\) is \(\alpha\,\delta t.\) How many punctures can I expect to get on a journey during which my total cycling time is \(T\)? When I get a puncture I stop immediately to repair it and the probability that, if I am repairing it at time \(t\), the repair will be completed in time \((t,t+\delta t)\) is \(\beta\,\delta t.\) If \(p(t)\) is the probability that I am repairing a puncture at time \(t\), write down an equation relating \(p(t)\) to \(p(t+\delta t)\), and derive from this a differential equation relating \(p'(t)\) and \(p(t).\) Show that \[ p(t)=\frac{\alpha}{\alpha+\beta}(1-\mathrm{e}^{-(\alpha+\beta)t}) \] satisfies this differential equation with the appropriate initial condition. Find an expression, involving \(\alpha,\beta\) and \(T\), for the time expected to be spent mending punctures during a journey of total time \(T\). Hence, or otherwise, show that, the fraction of the journey expected to be spent mending punctures is given approximately by \[ \quad\frac{\alpha T}{2}\quad\ \mbox{ if }(\alpha+\beta)T\text{ is small, } \] and by \[ \frac{\alpha}{\alpha+\beta}\quad\mbox{ if }(\alpha+\beta)T\text{ is large.} \]

1. In the differential equation \[ \frac{1}{y^{2}}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{1}{y}=\mathrm{e}^{2x} \] make the substitution \(u=1/y,\) and hence show that the general solution of the original equation is \[ y=\frac{1}{A\mathrm{e}^{x}-\mathrm{e}^{2x}}\,. \]
2. Use a similar method to solve the equation \[ \frac{1}{y^{3}}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{1}{y^{2}}=\mathrm{e}^{2x}. \]

Two identical particles of unit mass move under gravity in a medium for which the magnitude of the retarding force on a particle is \(k\) times its speed. The first particle is allowed to fall from rest at a point \(A\) whilst, at the same time, the second is projected upwards with speed \(u\) from a point \(B\) a positive distance \(d\) vertically above \(A\). Find their distance apart after a time \(t\) and show that this distance tends to the value \[ d+\frac{u}{k} \] as \(t\rightarrow\infty.\)

Find the two solutions of the differential equation \[ \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}=4y \] which pass through the point \((a,b^{2}),\) where \(b\neq0.\) Find two distinct points \((a_{1},1)\) and \((a_{2},1)\) such that one of the solutions through each of them also passes through the origin. Show that the graphs of these two solutions coincide and sketch their common graph, together with the other solutions through \((a_{1},1)\) and \((a_{2},1)\). Now sketch sufficient members of the family of solutions (for varying \(a\) and \(b\)) to indicate the general behaviour. Use your sketch to identify a common tangent, and comment briefly on its relevance to the differential equation.

A set of curves \(S_{1}\) is defined by the equation \[ y=\frac{x}{x-a}, \] where \(a\) is a constant which is different for different members of \(S_{1}.\) Sketch on the same axes the curves for which \(a=-2,-1,1\) and \(2\). A second of curves \(S_{2}\) is such that at each intersection between a member of \(S_{2}\) and a member of \(S_{1}\) the tangents of the intersecting curves are perpendicular. On the same axes as the already sketched members of \(S_{1},\) sketch the member of \(S_{2}\) that passes through the point \((1,-1)\). Obtain the first order differential equation for \(y\) satisfied at all points on all members of \(S_{1}\) (i.e. an equation connecting \(x,y\) and \(\mathrm{d}y/\mathrm{d}x\) which does not involve \(a\)). State the relationship between the values of \(\mathrm{d}y/\mathrm{d}x\) on two intersecting curves, one from \(S_{1}\) and one from \(S_{2},\) at their intersection. Hence show that the differential equation for the curves of \(S_{2}\) is \[ x=y(y-1)\dfrac{\mathrm{d}y}{\mathrm{d}x}. \] Find an equation for the member of \(S_{2}\) that you have sketched.

Let \(y,u,v,P\) and \(Q\) all be functions of \(x\). Show that the substitution \(y=uv\) in the differential equation \[ \frac{\mathrm{d}y}{\mathrm{d}x}+Py=Q \] leads to an equation for \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\) in terms of \(x,Q\) and \(u\), provided that \(u\) satisfies a suitable first order differential equation. Hence or otherwise solve \[ \frac{\mathrm{d}y}{\mathrm{d}x}-\frac{2y}{x+1}=\left(x+1\right)^{\frac{5}{2}}, \] given that \(y(1)=0\). For what set of values of \(x\) is the solution valid?

A damped system with feedback is modelled by the equation \[ \mathrm{f}'(t)+\mathrm{f}(t)-k\mathrm{f}(t-1)=0,\mbox{ }\tag{\(\dagger\)} \] where \(k\) is a given non-zero constant. Show that (non-zero) solutions for \(\mathrm{f}\) of the form \(\mathrm{f}(t)=A\mathrm{e}^{pt},\) where \(A\) and \(p\) are constants, are possible provided \(p\) satisfies \[ p+1=k\mathrm{e}^{-p}.\mbox{ }\tag{*} \] Show also, by means of a sketch, or otherwise, that equation \((*)\) can have \(0,1\) or \(2\) real roots, depending on the value of \(k\), and find the set of values of \(k\) for which such solutions of \((\dagger)\) exist. For what set of values of \(k\) do such solutions tend to zero as \(t\rightarrow+\infty\)?

Show Solution

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Solution: Suppose \(f(t) = Ae^{pt}\) is a solution, then \begin{align*} && 0 &= Ape^{pt} + Ae^{pt} - Ake^{p(t-1)} \\ \Leftrightarrow && 0 &= p +1 - ke^{-p} \\ \Leftrightarrow && p+1 &= ke^{-p} \end{align*}

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If we sketch \(y = x+1\) and \(y = ke^{-x}\) we can see that when \(k \geq 0\) there will be exactly one solution. If \(k < 0\) there can be no solutions (if \(k\) is large and negative) and two solutions if \(k\) is small and negative. There will be exactly one real root if \(y = x+1\) is tangent to \(y = ke^{-x}\). The gradient of \(y = ke^{-x}\) is \(-ke^{-x}\) so to have a gradient \(1\) we must have \(x+1 = -1 \Rightarrow x = -2\), but also \(x = -\ln(-k) \Rightarrow k = -e^{-2}\). Therefore there will be so solution as long as \(k \geq -e^{-2}\). The solutions will tend to \(0\) as \(t \to + \infty\) as long as the intersection point is less than zero. For \(k \geq 0\) they intersect at \(0\) when \(k =1\), so we would want \(k < 1\). All negative values of k will work since the intersection has to happen for negative \(p\). Therefore the range is \(-e^{-2} \leq k < 1\)

The normal to the curve \(y=\mathrm{f}(x)\) at the point \(P\) with coordinates \((x,\mathrm{f}(x))\) cuts the \(y\)-axis at the point \(Q\). Derive an expression in terms of \(x\), \(\mathrm{f}(x)\) and \(\mathrm{f}'(x)\) for the \(y\)-coordinate of \(Q\). If, for all \(x\), \(PQ=\sqrt{\mathrm{e}^{x^{2}}+x^{2}}\), find a differential equation satisfied by \(\mathrm{f}(x)\). If the curve also has a minimum point \((0,-2)\), find its equation.