Problems

Solution: Let \(v = \frac{\d x}{\d t}\) and notice that \(\frac{\d}{\d t} \left ( \frac{\d x}{\d t} \right) = \frac{\d }{\d x} \left ( v \right) \frac{\d x}{\d t} = v \frac{\d v}{\d x}\). Also notice that: \begin{align*} && v \frac{\d v}{\d x} &= 2x v \\ \Rightarrow && \frac{\d v}{\d x} &= 2x \\ \Rightarrow && v &= x^2 + C \\ \Rightarrow && \frac{\d x}{\d t} &= x^2 + C \\ \end{align*}

1. When \(t = 0, \frac{\d x}{\d t} = a^2\) so \(C = 0\), therefore \(\frac{\d x}{\d t} = x^2 \Rightarrow t = -x^{-1} + k\) and so \(k = a^{-1}\) and \(x = \frac{a}{1-at}\). As \(t\) increases from \(0\) the particle heads to infinity at an increasing rate, `reaching' infinity around \(t=\frac{1}{a}\)
2. When \(t = 0, \frac{\d x}{\d t} = a^2 + p\) so \(C = p\). Therefore \(\frac{\d x}{\d t} = x^2 + p \Rightarrow t = \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) + c\). When \(c = - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right)\), so \begin{align*} && t &= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{x}{\sqrt{p}} \right) - \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{a}{\sqrt{p}} \right) \\ &&&= \frac{1}{\sqrt{p}} \tan^{-1} \left ( \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} \right) \\ \Rightarrow && \frac{\sqrt{p}(x-a)}{\sqrt{p}-ax} &= \tan (\sqrt{p} t) \\ \Leftrightarrow && \sqrt{p}(x-a) &= \tan (\sqrt{p} t)(\sqrt{p}-ax) \\ \Leftrightarrow && x(\sqrt{p}+a\tan (\sqrt{p} t)) &= \sqrt{p} (\tan(\sqrt{p}t) + a) \\ \Leftrightarrow && x &= \frac{\sqrt{p} (\tan(\sqrt{p}t) + a)}{\sqrt{p}+a\tan (\sqrt{p} t)} \end{align*} The particle heads to \(\frac{\sqrt{p}}{a}\).
3. When \(t = 0, \frac{\d x}{\d t} = a^2-q^2\) so \(C = -q^2\). Therefore \begin{align*} && \frac{\d x}{\d t} &= x^2 -q^2 \\ \Rightarrow && \int \d t &= \int \frac{1}{(x-q)(x+q)} \d x \\ &&&= \frac{1}{2q} \int \left ( \frac{1}{x-q}- \frac{1}{x+q} \right )\d x \\ &&&= \frac{1}{2q} \left ( \ln (x-q) - \ln(x+q) \right) \\ &&&= \frac{1}{2q} \ln \left ( \frac{x-q}{x+q} \right)\\ \Rightarrow && \frac{x-q}{x+q} &= Ae^{2qt} \\ \underbrace{\Rightarrow}_{t= 0} && A &= \frac{a-q}{a+q} \\ \Rightarrow && x-q &= \frac{a-q}{a+q}e^{2qt}(x+q) \\ \Leftrightarrow && x\left (1-\frac{a-q}{a+q}e^{2qt} \right) &= q\left (1 + \frac{a-q}{a+q}e^{2qt} \right) \\ \Leftrightarrow && x &= q \frac{1 + \frac{a-q}{a+q}e^{2qt}}{1-\frac{a-q}{a+q}e^{2qt}} \end{align*}

Solution:

1. Since we have \(h'' + \omega^2 h = 0\) we must have that \(h(t) = A \cos \omega t + B \sin \omega t\). The initial conditions tell us that \(A = 0\) and \(B = L\), so \(h(t) = L \sin \omega t\).
   

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   Therefore we can see the angle at \(B\) is \(\omega t\) and so \(P\) has \(y\)-coordinate \((1-s)L \sin \omega t\) and \(x\)-coordinate \(sL \cos \omega t\)
2. If the position is \(\binom{sL \cos \omega t}{(1-s) L \sin \omega t}\) then the acceleration is \(-\omega^2 \binom{sL \cos \omega t}{(1-s) L \sin \omega t}\)
3. 

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   \begin{align*} \text{N2}(\rightarrow): && - F\cos \omega t + N \sin \omega t &= -m\omega^2 sL \cos \omega t\\ \text{N2}(\uparrow): && -mg + F\sin \omega t + N \cos \omega t &= -m\omega^2 (1-s) L \sin \omega t \\ \Rightarrow && \begin{pmatrix} \cos \omega t & -\sin \omega t \\ \sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} F \\ N \end{pmatrix} &= \begin{pmatrix} m\omega^2 s L \cos \omega t \\ mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\ \Rightarrow && \begin{pmatrix}F \\ N \end{pmatrix} &= \begin{pmatrix} \cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} m\omega^2 s L \cos \omega t \\ mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\ \Rightarrow && N &= m \omega^2 s L (-\sin \omega t \cos \omega t) + mg \cos \omega t - m \omega^2 (1-s)L \sin \omega t \cos \omega t \\ &&&=mg \cos \omega t - m \omega^2 L \sin \omega t \cos \omega t \\ &&&= mg \cos \omega t \left (1 - \frac{L \omega^2}{g} \sin \omega t \right) \\ &&&= mg (1 - k \sin \omega t) \cos \omega t \\ \Rightarrow && F &= m \omega^2 s L \cos^2 \omega t + mg \sin \omega t - m \omega^2 (1-s) L \sin ^2 \omega t \\ &&&= m \omega^2 s L + mg \sin \omega t - m \omega^2 L \sin^2 \omega t \\ &&&= mg \frac{\omega^2 L}{g} s + mg(1-\frac{\omega^2 L}{g} \sin \omega t)\sin \omega t \\ &&&= mg sk + mg(1-k \sin \omega t) \cos \omega t \tan \omega t \\ &&&= mgsk + N \tan \omega t \end{align*}
4. The particle will not slip if \(F < \tan \alpha N\). When \(t = 0\), \(N = mg, F = mgsk\), but clearly \(sk < \tan \alpha \Rightarrow mgsk = F < \tan \alpha mg = \tan \alpha N\). The particle will slip when: \(F > \tan \alpha N\), but we have \(F = mgsk + N \tan \omega t\). Clearly when \(\omega t = \alpha\) we have reached a point where \(F > \tan \alpha N\). Therefore we must slip before we reach this point, ie at a point where the plank makes an angle of less than \(\alpha\) to the horizontal. Notice also that \(N\) changes sign when \(1-k \sin \omega t = 0\), however, to do this \(N\) must become very small, smaller than \(mgsk\), therefore we must slip before this point too. Since we slip before either condition occurs, we must be in a position when \(N\) is positive AND the plank still makes a shallow angle.

Solution:

1. Looking for solution of the form \(y = mx+c\) we find that \(m = mx+c+x+1 \Rightarrow m = -1, c = -2\). At stationary points \(\frac{\d y}{\d x} = 0 \Rightarrow y = -x-1\). If \(y_3(0)= k < -2\) then the solution curve lies below \(y = -x-2\) and therefore it cannot cross \(y = -x -2\) to reach \(y = -x-1\) for a stationary point. Suppose \(Y = y+x\) then \(\frac{\d Y}{\d x} = \frac{\d y}{\d x} + 1=Y+2 \Rightarrow Y = Ae^x-2 \Rightarrow y= (k+2)e^x-x-2\)
   

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2. \(\,\) \begin{align*} && m &= x^2 + (mx+c)^2 -2x(mx+c) - 4x+4(mx+c) + 3 \\ &&0&= (m^2-2m+1)x^2+(2mc-2c-4+4m)x + (c^2+4c+3-m)\\ \Rightarrow && m &= 1 \\ \Rightarrow && 0 &= c^2+4c+2 \\ \Rightarrow &&&= (c+2)^2-2 \\ \Rightarrow && c &= -2 \pm \sqrt{2} \end{align*} Therefore the lines are \(y = x -2-\sqrt{2}\) and \(y = x -2+\sqrt{2}\). Any stationary point will satisfy \(y' = 0\), ie \(0 = x^2+y^2-2xy-4x+4y+3 = (x-y)^2-4(x-y)+3 = (x-y-3)(x-y-1)\) therefore they must lie on \(y = x-1\) or \(y = x-3\). Any point between these lines must have negative gradient (since one factor is positive and one factor is negative).
   

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Solution:

1. Let \(k = 1\), then \begin{align*} \dot{x} &= - x - y \\ \dot{y} &= x - y \\ \dot{x}-\dot{y} &= -2x \\ \ddot{x} &= -\dot{x}-\dot{y} \\ &= -\dot{x} - (\dot{x}+2x) \\ &= -2\dot{x}- 2x \\ \dot{x}+\dot{y} &= -2y \\ \ddot{y} &= \dot{x}-\dot{y} \\ &= -2y-2\dot{y} \end{align*} So we have an auxiliary equation for \(x\) and \(y\) which is \(\lambda^2 + 2 \lambda+2 = 0 \Rightarrow \lambda = -1 \pm i\). Therefore \(x = Ae^{-t} \cos t + B e^{-t} \sin t, y = Ce^{-t}\cos t + De^{-t} \sin t\). We also must have that, \(A = 1, C = 0\), so \(x = e^{-t} \cos t + Be^{-t} \sin t\) and \(y = De^{-t} \sin x\). \begin{align*} \dot{y} &= -De^{-t} \sin t +De^{-t} \cos t \\ &= e^{-t} \cos x + Be^{-t} \sin t- De^{-t} \sin t \\ \end{align*} therefore \(B = 0, D = 1\) and \(x = e^{-t} \cos t, y = e^{-t} \sin t\)
   

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   \begin{align*} y &= e^{-t} \sin t \\ \dot{y} &= -e^{-t} \sin t + e^{-t} \cos t \\ \dot{x} &= e^{-t} \cos t -e^{-t} \sin t \end{align*}
   

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2. \begin{align*} \dot{x} = -x \\ \dot{y} = x-y \end{align*} So \(x = e^{-t}\). \(\dot{y} + y = e^{-t}\) so \(y = (t+B)e^{-t}\) and so \(y =te^{-t}\).
   

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Solution:

1. \(\,\) \begin{align*} && f(0+x) &= f(0)f(x) \\ \Rightarrow && f(0) &= 0, 1\\ &&\text{since }f'(0) \neq 0 & \text{ there is some non-zero } f(x) \\ \Rightarrow && f(0) &= 1 \end{align*} \begin{align*} && f'(x) &= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ &&&= \lim_{h\to 0} \frac{f(x)f(h)-f(x)}{h} \\ &&&= f(x) \cdot \lim_{h\to 0} \frac{f(h)-1}{h} \\ &&&= f(x) \cdot \lim_{h\to 0} \frac{f(0+h)-f(0)}{h} \\ &&&= f(x) \cdot f'(0) \\ &&&= kf(x) \end{align*} Since \(f'(x) = kf(x)\) we must have \(\frac{f'(x)}{f(x)} = k \Rightarrow \ln f(x) = kx + c \Rightarrow f(x) = Ae^{kx}\) but \(f(0) = 1\) so \(f(x) = e^{kx}\)
2. Consider \begin{align*} && g(0+0) &= \frac{g(0)+g(0)}{1+(g(0))^2} \\ \Rightarrow && g(0)(1+(g(0))^2)&= 2g(0) \\ \Rightarrow && 0 &= g(0)\left (1- (g(0))^2 \right) \\ \Rightarrow && g(0) &= -1, 0, 1 \\ \Rightarrow && g(0) &= 0 \tag{\(|g(0)| < 1\)} \end{align*} \begin{align*} && g'(x) &=\lim_{h\to 0} \frac{g(x+h)-g(x)}{h} \\ &&&= \lim_{h\to 0} \frac{\frac{g(x)+g(h)}{1+g(x)g(h)}-g(x)}{h} \\ &&&= \lim_{h\to 0} \frac{g(x)+g(h)-g(x)(1+g(x)g(h))}{h(1+g(x)g(h))} \\ &&&= \lim_{h\to 0} \frac{g(h)-g(x)(g(x)g(h))}{h(1+g(x)g(h))} \\ &&&= (1-(g(x))^2) \cdot \lim_{h \to 0} \frac{1}{1+g(x)g(h)} \cdot \lim_{h \to 0} \frac{g(h)}{h} \\ &&&= (1-(g(x))^2) \cdot \frac{1}{1+g(x)\cdot 0} \cdot \lim_{h \to 0} \frac{g(h) - g(0)}{h} \\ &&&= (1-(g(x))^2) \cdot g'(0)\\ &&&= k (1-(g(x))^2) \\ \end{align*} Let \(y = g(x)\) so \begin{align*} && y' &= k(1-y^2) \\ \Rightarrow && kx &= \int \frac{1}{1-y^2} \d y \\ \Rightarrow &&&= \int \frac12\left ( \frac{1}{1-y} + \frac{1}{1+y} \right) \d y \\ &&&= \frac12\ln \left ( \frac{1+y}{1-y} \right) + C \\ x = 0, y = 0: && 0 &= \ln 1 + C \\ \Rightarrow && C &= 0 \\ \Rightarrow && \frac{1+y}{1-y} &= e^{2kx} \\ \Rightarrow && 1+y &= e^{2kx} - e^{2kx}y \\ \Rightarrow && y &= \frac{e^{2kx}-1}{e^{2kx}+1} \\ &&&= \tanh kx \end{align*}

Solution: \begin{questionparts} \item \begin{align*} && \dfrac{\d }{\d x} \left( \s(x)^3 + \c(x)^3 \right) &= 3\s(x)^2\s'(x) + 3\c(x)^2 \c'(x) \\ &&&= 3\s(x)^2\c(x)^2 - 3\c(x)^2\s(x)^2 \\ &&&= 0 \\ \\ \Rightarrow && \s(x)^3 + \c(x)^3 &= \text{constant} \\ &&&= \s(0)^3 + \c(0)^3 \\ &&&= 1 \end{align*} \item \begin{align*} \frac{\d }{\d x} \, \Big( \s(x) \c(x) \Big) &= \s'(x) \c(x) + \s(x)\c'(x) \\ &= \c(x)^3 - \s(x)^3 \\ &= \c(x)^3 - (1-\c(x)^3) \\ &= 2\c(x)^3 - 1 \\ \\ \dfrac{\d }{\d x} \left( \dfrac{\s(x)}{\c(x)} \right) &= \frac{\s'(x)\c(x) - \s(x)\c'(x)}{\c(x)^2} \\ &= \frac{\c(x)^3 + \s(x)^3}{\c(x)^2} \\ &= \frac{1}{\c(x)^2} \\ \end{align*} \item \begin{align*} \int \s(x)^2 \d x &= -\int -\s(x)^2 \d x \\ &= -\int \c'(x) \d x \\ &= - \s(x) +C \\ \\ \int \s(x)^5 \, \d x &= \int \s(x)^2 \s(x)^3 \d x \\ &= \int \s(x)^2 (1 - \c(x)^3) \d x \\ &= -\int \c'(x) (1 - \c(x)^3) \d x \\ &= - c(x) + \frac{\c(x)^4}{4} + C \end{align*} \item If \(u = \s(x), \frac{\d u}{\d x} = \c(x)^2\) \begin{align*} \int \frac{1}{(1-u^3)^{\frac{2}{3}}} \, \d u &= \int \frac{1}{(1-\s(x)^3)^{\frac{2}{3}}} \c(x)^2 \d x \\ &= \int 1 \d x \\ &= x + C \\ &= \s^{-1}(u) + C \\ \\ \int \frac{1}{{(1-u^3)^{\frac{4}{3}}}} \d u &= \int \frac1{(1-\s(x)^3)^{\frac43} }\c(x)^2 \d x \\ &= \int \frac1{(\c(x)^3)^{\frac43}} \c(x)^2 \d x \\ &= \int \frac1{\c(x)^2} \d x \\ &= \frac{\s(x)}{\c(x)} + C \\ &= \frac{u}{(1-u^3)^{\frac13}} + C \\ \end{align*} \begin{align*} && \int {(1-u^3)}^{\frac{1}{3}} \, \d u &= \int (1-s(x)^3)^{\frac13} c(x)^2 \d x \\ &&&= \int \c(x)^3 \d x = I\\ &&&= \int \c(x) s'(x) \d x \\ &&&= \left [\c(x) \s(x) \right] + \int \s(x)^2 s(x) \d x \\ &&&= \c(x) \s(x) + \int (1 - \c(x)^3) \d x + C \\ &&&= \c(x) \s(x) + x - I + C \\ \Rightarrow && I &= \frac{x + \c(x) \s(x)}{2} + k \\ \Rightarrow && &= \frac12 \l \s^{-1}(u) + u \sqrt[3](1-u^3)\r + k \end{align*}

Solution:

1. Suppose \(v = \sqrt{y} \Rightarrow v^2 = y \Rightarrow 2v v' = y'\) so \begin{align*} && 2vv' &= \alpha v-\beta v^2 \\ \Rightarrow && 2v' &= \alpha - \beta v \\ \Rightarrow && v' + \frac{\beta}{2} v &= \frac{\alpha}{2} \\ \Rightarrow && \frac{\d}{\d t} \left (e^{\beta t/2} v \right) &= \frac{\alpha}{2} e^{\beta t/2} \\ \Rightarrow && e^{\beta t/2} v &=C+ \frac{\alpha}{\beta}e^{\beta t /2} \\ \Rightarrow && v &= Ce^{-\beta t/2} + \frac{\alpha}{\beta} \\ \Rightarrow && \sqrt{y} &= Ce^{-\beta t/2} + \frac{\alpha}{\beta} \\ y(0) = 0: && 0 &= C+\frac{\alpha}{\beta} \\ \Rightarrow && \sqrt{y} &= \frac{\alpha}{\beta} \left (1-e^{-\beta t/2} \right) \\ \Rightarrow && y &= \frac{ \alpha^2}{\beta^2} \left (1-e^{-\beta t/2} \right)^2 \end{align*}
2. Try \(v = y^{1/3} \Rightarrow v^3 = y \Rightarrow 3v^2 v' = y'\) so \begin{align*} && y' &= \alpha v^2 - \beta y \\ \Rightarrow && 3v^2v' &= \alpha v^2 - \beta v^3 \\ \Rightarrow && v' +\frac{\beta}{3} v &= \frac{\alpha}{3} \\ \Rightarrow && (v e^{\beta t/3})' &= \frac{\alpha}{3}e^{\beta t/3} \\ \Rightarrow && v &= Ce^{-\beta t/3} + \frac{\alpha}{\beta} \\ v(0) = 0: && v &= \frac{\alpha}{\beta} \left (1 - e^{-\beta t/3} \right) \\ \Rightarrow && y &= \frac{\alpha^3}{\beta^3} \left (1 - e^{-\beta t/3} \right) ^3 \end{align*}
3. \(y_1 = (1-e^{-\beta t/2})^2, y_2 = (1-e^{-\beta t/3})^3\)
   

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   By considering the differential equation, notice that \(0 < y_i < 1\) so \(y^{1/2} > y^{2/3}\) and therefore \(y_1' > y_2'\) and so \(y_1\) increases faster.

Solution: Consider $x^a \big(x^b (x^cy)'\big)'$ then \begin{align*} x^a \big(x^b (x^cy)'\big)' &= x^a \big (bx^{b-1}(x^c y)'+x^b(x^cy)'' \big ) \\ &= x^a \big (bx^{b-1} (cx^{c-1}y + x^c y') + x^b(c(c-1)x^{c-2}y + 2cx^{c-1}y' + x^cy'') \\ &= x^{a+b+c}y'' + (2cx^{c-1+b+a}+bx^{c+b-1+a})y'+(c(b+c-1))x^{a+b+c-2} y \end{align*} So we need: \begin{align*} &&& \begin{cases} a+b+c &= 2 \\ 2c+b &= 1-2p \\ c(b+c-1) &= p^2-q^2 \end{cases} \\ \Rightarrow && c((1-2p)-2c+c-1) &=p^2-q^2 \\ \Rightarrow && c^2+2pc &= q^2-p^2 \end{align*}

Solution: \((x, y) = (f(\theta)\cos(\theta), f(\theta)\sin(\theta))\) so \begin{align*} \frac{dy}{d\theta} &= -f(\theta)\sin(\theta) + f'(\theta)\cos(\theta) \\ \frac{dx}{d\theta} &= f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) \\ \frac{dy}{dx} &= \frac{-f(\theta)\sin(\theta) + f'(\theta)\cos(\theta)}{f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) } \\ &= \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \end{align*} If the curves meet at right angles then the product of their gradients is \(-1\), ie \begin{align*} \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \cdot \frac{-g(\theta)\tan(\theta) + g'(\theta)}{g(\theta) + g'(\theta)\tan(\theta) } &= -1 \\ f(\theta)g(\theta)\tan^2 \theta - f(\theta)g'(\theta)\tan \theta - f'(\theta)g(\theta)\tan \theta + f'(\theta)g'(\theta) &= \\ \quad - \l f(\theta)g(\theta) + f(\theta)g'(\theta)\tan(\theta) + f'(\theta)g(\theta)\tan(\theta) + f'(\theta)g'(\theta)\tan^2 \theta \r \\ \tan^2\theta \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r + f'(\theta)g'(\theta) + f(\theta)g(\theta) &= 0 \\ (\tan^2\theta + 1) \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r &= 0 \\ f(\theta)g(\theta) + f'(\theta)g'(\theta) &= 0 \end{align*} \(g(\theta) = a(1+\sin\theta), g'(\theta) = a\cos\theta\) Therefore \(f'(\theta)a\cos \theta+f(\theta)a(1+\sin(\theta)) = 0\) \begin{align*} && \frac{f'(\theta)}{f(\theta)} &= -\sec(\theta) - \tan(\theta) \\ \Rightarrow && \ln(f(\theta)) &= -\ln |\tan(\theta) + \sec(\theta)| + \ln |\cos(\theta)| + C \\ \Rightarrow && f(\theta) &= A \frac{\cos \theta}{\tan \theta + \sec \theta} \\ &&&= A \frac{\cos^2 \theta}{\sin \theta + 1} \\ &&&= A \frac{1-\sin^2 \theta}{\sin \theta + 1} \\ &&&= A (1-\sin \theta) \end{align*} When \(\theta = -\frac12 \pi, r = 4\), so \(A = 2\).

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Solution:

1. When \(k =1\), we have \((1-x^2)(y')^2 + y^2 = 1\). Notice that if \(y_1 = x\) we have \(y_1' = 1\) and \((1-x^2) \cdot 1 + x^2 = 1\) so \(y_1\) is a solution, and we are allowed to assume this is the only solution. And notice that \(y_1(1) = 1\).
2. When \(k = 2\) we have \((1-x^2)(y')^2 + 4y^2 = 4\). Trying \(y_2 = 2x^2-1\) we see that \(y_2' = 4x\) and \((1-x^2)(4x)^2 + 4(2x^2-1)^2 = 16x^2-16x^4+16x^4-16x^2+4 = 4\). We can also check that \(y(1) = 2 \cdot 1^2 - 1 = 1\)
3. Let \(z(x) = 2(y_n(x))^2-1\), then \begin{align*} && \frac{\d z}{\d x} &= 4y'_n(x)y_n(x) \\ \Rightarrow && LHS &= (1-x^2)\left ( \frac{\d z}{\d x} \right)^2 + 4n^2 z^2 \\ &&&= (1-x^2)16(y'_n(x))^2(y_n(x))^2 + 4n^2(2(y_n(x))^2-1)^2 \\ &&&= 16y_n^2(1-x^2) \left [\frac{n^2-n^2y_n^2}{(1-x^2)} \right] + 16n^2y_n^4-16n^2y_n^2+4n^2 \\ &&&= 4n^2 = RHS \end{align*} Therefore \(y_{2n}(x) = 2(y_n(x))^2-1 = y_2(y_n(x))\) (notice also that \(z(1) = 2(y_n(1))^2-1 = 2-1 = 1\)).
4. Let \(v(x) = y_n(y_m(x))\) so \begin{align*} && (y_m')^2 &= \frac{m^2(1-y_m^2)}{1-x^2} \\ && (y_n')^2 &= \frac{n^2(1-y_n^2)}{1-x^2} \\ \\ && \frac{\d v}{\d x} &= y_n'(y_m(x)) \cdot y_m'(x) \\ && (1-x^2)(v')^2 &= (1-x^2) \cdot (y_n'(y_m(x)))^2 (y_m')^2 \\ &&&= (1-x^2) \cdot (y_n'(y_m(x)))^2 \left ( \frac{m^2(1-y_m^2)}{1-x^2} \right) \\ &&&= (y_n'(y_m(x)))^2 m^2(1-y_m^2) \\ &&&= \frac{n^2(1-(y_n(y_m(x)))^2)}{1-y_m^2}m^2(1-y_m^2) \\ &&&= n^2m^2(1-v^2) \end{align*} Therefore \(v\) satisfies our differential equation and \(v(1) = y_n(y_m(1)) = y_n(1) = 1\) so it must be our desired solution.

Solution:

1. \begin{align*} && (y+x)\frac{\d y}{\d x} &= y-x \\ \Rightarrow && (r \sin \theta + r \cos\theta) \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} &= (r \sin\theta - r \cos\theta) \\ \Rightarrow && ( \sin \theta + \cos\theta) \frac{dy}{d\theta} &= (\sin\theta - \cos\theta){\frac{dx}{d\theta}} \\ \Rightarrow && ( \sin \theta + \cos\theta) \left ( \frac{dr}{d\theta} \cos \theta - r \sin \theta\right ) &= (\sin\theta - \cos\theta)\left ( \frac{dr}{d\theta} \sin\theta + r \cos \theta\right) \\ \Rightarrow && \frac{dr}{d\theta} \left (\sin \theta \cos \theta + \cos^2 \theta - \sin^2 \theta + \sin \theta \cos \theta \right)&= r \left (\sin \theta \cos \theta - \cos^2 \theta + \sin^2 \theta + \sin\theta \cos \theta\right) \\ \Rightarrow && \frac{dr}{d\theta}&= -r \\ \end{align*} Therefore \(r = Ae^{-\theta}\)
   

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2. \begin{align*} && \left( y+x -x(x^2+y^2) \right) \, \frac{\d y }{\d x} &= y-x - y(x^2+y^2) \\ \Rightarrow && \left( r \sin \theta+r\cos \theta -r^3\cos \theta \right) \, \frac{\d y }{\d \theta} &= \left ( r \sin \theta- r \cos \theta- r^3\sin \theta \right)\frac{\d x }{\d \theta} \\ \Rightarrow && \left( r \sin \theta+r\cos \theta -r^3\cos \theta \right) \, \left (\frac{\d r}{\d \theta} \sin \theta + r \cos \theta \right) &= \\ && \left ( r \sin \theta- r \cos \theta- r^3\sin \theta \right)&\left (\frac{\d r}{\d \theta} \cos \theta - r \sin \theta \right) \\ \Rightarrow && \frac{\d r}{\d \theta} \left (\sin \theta ( \sin \theta + \cos \theta - r^2 \cos \theta) - \cos \theta (\sin \theta - \cos \theta - r^2 \sin \theta) \right) &= \\ && r ( -\sin \theta (\sin \theta - \cos \theta - r^2 \sin \theta) - \cos \theta ( \sin \theta + \cos \theta &- r^2 \cos \theta)) \\ \Rightarrow && \frac{\d r}{\d \theta} &= r ( -1 +r^2) \\ \Rightarrow && \int \frac{1}{r(r-1)(r+1)} \d r &= \int \d \theta \\ \Rightarrow && \int \l \frac{-1}{r} + \frac{1}{2(r-1)} + \frac{1}{2(r+1)} \r \d r &= \int \d \theta \\ \Rightarrow && \l -\log r+ \frac12 \log (1+r)+ \frac12 \log (1-r)\r + C &= \theta \\ \Rightarrow && \frac12 \log \left (\frac{1-r^2}{r^2} \right) + C &= \theta \\ \Rightarrow && \log \left (\frac{1}{r^2}-1 \right) + C &= 2\theta \\ \Rightarrow && r &= \frac{1}{1 + Ae^{2\theta}} \\ \end{align*}
   

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Solution:

1. \(\,\)
   

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   Note that we have a right angled triangle, with the sides in a ratio of \(m\). So if our target length is \(x\) we have \(x^2 + (mx)^2 = c^2\) and so \(x = c(m^2+1)^{-\frac12}\)
2. The distance from the origin to \(L\) is \(a = c(m^2+1)^{-\frac12}\) so \begin{align*} && a^2(m^2+1) &= c^2 \\ && \frac{c-y(x)}{0-x} &= y' \\ \Rightarrow && c-y &= -xy' \\ \Rightarrow && a^2((y')^2+1) &= (y-xy')^2 \\ \\ && 2a^2y'y'' &= 2(y-xy')(y'-xy''-y') \\ &&&= 2(xy'-y)xy'' \\ \Rightarrow && y'' &= 0 \\ \text{ or } && 2a^2y' &= 2(xy'-y)x \end{align*} If \(y'' = 0\) then \(y = mx + c\) and the result follows immediately. \begin{align*} && 0 &= (a^2-x^2)y' + yx \\ \Rightarrow &&\frac1{y} y' &= -\frac{x}{a^2-x^2} \\ \Rightarrow && \ln y &= \frac12\ln (a^2-x^2) + K \\ \Rightarrow && y^2 &= M(a^2-x^2) \\ \Rightarrow && x^2 + y^2 &= a^2 \end{align*} Where in the last step we know the tangents from an ellipse are not all equidistant to the origin.
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Solution: \begin{align*} \frac{\d y}{\d x} &= \frac{\d }{\d x} \l y \r \\ &= \frac{\d }{\d x} \l xu \r \\ &\underbrace{=}_{\text{product rule}} \frac{\d}{\d x} \l x \r u + x \frac{\d}{\d x} \l u \r \\ &= u + x \frac{\d u}{\d x} \end{align*} \begin{questionparts} \item \begin{align*} && \frac{\d y}{\d x} &= \frac{2y + x}{y - 2x} \\ && u + x \frac{\d u}{\d x} &= \frac{2u + 1}{u - 2} \\ && x \frac{\d u}{\d x} &= \frac{2u-1-u^2+2u}{u-2} \\ \Rightarrow && \int \frac{2-u}{u^2-4u+1} \d u &= \int \frac{1}{x} \d x \\ && \int \frac{2-u}{(u-2)^2-5} \d u &= \int \frac1x \d x \\ && -\frac12\ln| (u-2)^2 - 5| &= \ln x + C \\ (x,y) = (1,1): && - \ln 2 &= C \\ \Rightarrow && \ln x^2 &= \ln 4 - \ln |5 - (u-2)^2| \\ \Rightarrow && x^2 &= \frac{4}{5- (u-2)^2} \\ \Rightarrow && 4 & = x^2(5 - (\frac{y}{x} - 2)^2) \\ &&&= 5x^2 - (y-2x)^2 \\ &&&= x^2+4xy-y^2 \end{align*} \item It would be convienient if \(x-2y -4 = X-2Y\) and \(2x+y-3 = 2X+Y\), ie \(a-2b = 4\) and \(2a+b = 3\), ie \(a = 2, b = -1\). Now our differential equation is: \begin{align*} && \frac{\d Y}{\d X} &= \frac{X - 2Y}{2X+Y} \\ && \frac{\d X}{\d Y} &= \frac{2X + Y}{X-2Y} \end{align*} This is the same differential equation we have already solved, just with the roles of \(x\) and \(y\) interchanged with \(Y\) and \(X\) and with the point \((0,3)\) being on the curve, ie: \(Y^2 + 4XY-X^2 = c\) and \(c = 9\), therefore our equation is: \[ (y-1)^2 + 4(y-1)(x+2)-(x+2)^2 = 9\]

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(T > t + \delta t | T > t) &= \frac{\mathbb{P}(T < t + \delta t)}{\mathbb{P}(T > t )} \\ &&&= \frac{\int_t^{t+\delta t} f(s) \d s}{1-F(t)} \\ &&&\approx \frac{f(t)\delta t}{1-F(t)} \\ &&&= h(t) \delta t \end{align*}
2. If \(F(t) = t/a\) then \(f(t) = 1/a\) and \(h(t) = \frac{1/a}{1-t/a} = \frac{1}{a-t}\)
   

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3. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \frac{1}{t} \\ \Rightarrow && -\ln (1-F) &= \ln t + C\\ \Rightarrow && 1-F &= \frac{A}{t} \\ && F &= 1 - \frac{A}{t} \\ F(a) = 0: && F &= 1 - \frac{a}{t} \\ && f(t) &= \frac{a}{t^2} \end{align*}
4. (\(\Rightarrow\)) \begin{align*} && \frac{F'}{1-F} &= k \\ \Rightarrow && -\ln(1-F) &= kt+C \\ \Rightarrow && 1-F &= Ae^{-kt} \\ F(b) = 0: && 1 &= Ae^{-kb} \\ \Rightarrow && 1-F &= e^{-k(t-b)}\\ \Rightarrow && f &= ke^{-k(t-b)} \\ \end{align*} (\(\Leftarrow\)) \(f(t) = ke^{-k(t-b)} \Rightarrow F(t) = 1-e^{-k(t-b)}\) and the result is clear.
5. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \left ( \frac{\lambda}{\theta^{\lambda}} \right) t^{\lambda-1} \\ \Rightarrow && -\ln(1-F) &= \left ( \frac{t}{\theta} \right)^{\lambda} +C\\ \Rightarrow && F &= 1-A\exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ F(0) = 0: && 0 &= 1-A \\ \Rightarrow && F &= 1 - \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ \Rightarrow && f &= \lambda t^{\lambda -1} \theta^{-\lambda} \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \end{align*}

Solution:

1. \begin{align*} && I - I_1 &= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) \d x - \int_0^1 \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) - \left ( y' + y \tan x \right)^2 \d x\\ &&&= \int_0^1 \left (-y^2-2yy' \tan x - y^2 \tan^2 x \right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2(1+ \tan^2 x )\right) \d x\\ &&&= \int_0^1 \left (-2yy' \tan x - y^2 \sec^2 x\right) \d x\\ &&&= \int_0^1 -\frac{\d}{\d x} \left (y^2 \tan x \right) \d x\\ &&&= \left [-y^2 \tan x \right]_0^1 \\ &&&= 0 \\ \\ \Rightarrow && I &= I_1 = \int_0^1 \left ( y' + y \tan x \right)^2 d x \geq 0 \end{align*} The only way \(I_0 = 0\) is is \(y' + y \tan x =0\), so \begin{align*} && \frac{\d y}{\d x} &= - y \tan x \\ \Rightarrow && \int \frac{1}{y} &= \int -\tan x \d x \\ \Rightarrow && \ln |y| &= \ln |\cos x| + C \\ \Rightarrow && y &= A \cos x \\ \Rightarrow && A &= 0 \Rightarrow y = 0 \end{align*}
2. Let \(J_1 = \int_0^1 (y'+ay\tan ax)^2 \d x\), then \begin{align*} && J-J_1 &= \int_0^1 \left ( \left ( y' \right)^2 - a^2y^2 \right) - \left ( y' + ya \tan ax \right)^2 \d x\\ &&&= \int_0^1 \left (-a^2y^2-2yy' a \tan a x-y^2a^2 \tan^2 ax \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2(1+\tan^2 ax) \right) \d x \\ &&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2\sec^2 ax \right) \d x \\ &&&= \left [ - a y^2 \tan a x \right]_0^1 = 0 \end{align*} This is true if \(a < \frac{\pi}{2}\), since otherwise we might care about the order of the zero for \(y\) at \(x = 1\). Consider \(y = \cos \frac{\pi}{2} x\), then \(y' = -\frac{\pi}{2} \sin^2\frac{\pi}{2} x\) and \begin{align*} && \int_0^1 \frac{\pi^2}{4} \left (\sin^2 \frac{\pi}{2}x - \cos^2 \frac{\pi}{2} x \right) \d x &= -\frac{\pi^2}{4} \int_0^1 \cos(\pi x) \d x \\ &&&= 0 \end{align*}