Problems

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. 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: Points always maintain a constant fraction of the distance from the start, so the point distance \(x\) from the start at time \(t\) is moving with speed \(\frac{x}{a+ut} u\) The point is moving with speed \(v+\frac{x}{a+ut} u\) or in other words \begin{align*} && \frac{\d x}{\d t} &= v + \frac{x}{a+ut}u \\ \Rightarrow && \frac{\d x }{\d t} - \frac{u}{a+ut} x &= v \\ \Rightarrow && \frac{1}{a+ut} \frac{\d x}{\d t} - \frac{u}{(a+ut)^2} x &= \frac{1}{a+ut} v\\ \Rightarrow && \frac{\d}{\d x} \left ( \frac{x}{a+ut} \right) &= \frac{v}{a+ut} \\ \Rightarrow && \frac{x}{a+ut} &=\frac{v}{u} \ln (a + ut) + C \\ t = 0, x = 0: && 0 &= \frac{v}{u} \ln a + C \\ \Rightarrow && x &= (a+ut) \frac{v}{u} \ln \left ( \frac{a+ut}{a} \right) \\ \\ \Rightarrow && 1 &= \frac{v}{u} \ln \left ( \frac{a+uT}{a} \right) \\ \Rightarrow && e^k &= 1 + \frac{uT}{a} \\ \Rightarrow && uT &= a(e^k-1) \end{align*} On the return journey, we have \begin{align*} && \frac{\d x}{\d t} &= \frac{x}{a+ut}u - v \\ \Rightarrow && \frac{\d x}{\d t} - \frac{u}{a+ut} x &= - v \\ \Rightarrow && \frac{\d }{\d x} \left ( \frac{x}{a+ut} \right) &= -\frac{v}{a+ut} \\ \Rightarrow &&f &= -\frac{v}{u} \ln(a+ut) + K \\ t = T, f = 1: && 1 &= -\frac{v}{u}\ln(a+uT) + K \\ \Rightarrow && f &= 1+\frac{v}{u}\ln \left ( \frac{a+uT}{a+ut} \right) \\ \Rightarrow && 0 &= 1+\frac{v}{u} \ln \left ( \frac{a+uT}{a+uT_2} \right) \\ \Rightarrow && e^k &= \frac{a+uT_2}{a+uT}\\ \Rightarrow && uT_2 &= (a+uT)e^k - a \\ \Rightarrow && T_2 - T &= \frac{1}{u} \left ( (a+uT)e^k - a - uT\right) \\ &&&= \frac{1}{u} \left ((a+a(e^k-1))e^k-a-a(e^k-1) \right) \\ &&&= \frac{1}{u} \left (ae^{2k} -ae^k \right) \\ &&&= \frac{ae^k}{u} \left ( e^k-1 \right) \\ &&&= Te^k \end{align*}

Solution:

1. \(\,\) \begin{align*} && y &= xv \\ && y' &= v + xv' \\ \Rightarrow && 0 &= x^2 v \left ( v + x\frac{\d v}{\d x} \right) +(x^2v^2) - 2x^2 \\ &&&= 2x^2v^2 + x^3 v \frac{\d v}{\d x} - 2x^2 \\ \Rightarrow && 0 &= xv \frac{\d v}{\d x} + 2v^2-2 \\ \\ \Rightarrow && \frac{v}{1-v^2} \frac{\d v}{\d x} &= \frac{2}{x} \\ \Rightarrow && \int \frac{v}{1-v^2} \d v &=2 \ln |x| \\ \Rightarrow && -\frac12\ln |1-v^2| &= 2\ln |x| + C \\ \Rightarrow && 4\ln |x| + \ln |1-v^2| &= K \\ \Rightarrow && x^4(1-v^2) &= K \\ \Rightarrow && x^2(x^2-y^2) &= K \end{align*}
2. \(\,\) \begin{align*} && 0 &= xv \left (v +x \frac{\d v}{\d x} \right) + 6x + 5xv \\ &&&= x^2 v \frac{\d v}{\d x} +xv^2 + 6x+5xv \\ \Rightarrow && 0 &= xv\frac{\d v}{\d x} +v^2 +5v+6 \\ \Rightarrow && -\int \frac{1}{x} \d x &=\int \frac{v}{v^2+5v+6} \d v \\ \Rightarrow && -\ln |x| &= \int \frac{v}{(v+2)(v+3)} \d v \\ &&&= \int \left (\frac{3}{v+3} - \frac{2}{v+2} \right) \d v \\ \Rightarrow && -\ln |x| &= 3\ln |v+3| - 2 \ln |v+2| + C\\ \Rightarrow && -\ln |x| &= \ln \frac{|v+3|^3}{|v+2|^2} + C \\ \Rightarrow && \frac{1}{|x|}|v+2|^2 &= A|v+3|^3 \\ \Rightarrow && \frac{1}{|x|}|\frac{y}{x} + 2|^2&= A|\frac{y}{x} + 3|^3 \\ \Rightarrow && \frac{1}{|x|^3} |y +2x|^2 &= \frac{A}{|x|^3}|y + 3x|^3 \\ \Rightarrow && (y+2x)^2 &= A|y+3x|^3 \end{align*}

Solution:

1. \begin{align*} && 0 &= \frac{\d u}{\d x} - \left ( \frac{x+2}{x+1} \right)u \\ \Rightarrow && \int \frac1u \d u &= \int 1 + \frac1{x+1} \d x \\ \Rightarrow && \ln |u| &= x + \ln |x+1| + C \\ \Rightarrow && u &= A(x+1)e^x \end{align*}
2. If \(y = ze^{-x}\), \(y' = (z'-z)e^{-x}\), \(y'' = (z''-2z'+z)e^{-x}\) \begin{align*} && 0 &= (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y \\ y = ze^{-x}: && 0 &= (x+1) \left ( \frac{\d^2 z}{\d x^2} - 2\frac{\d z}{\d x} +z\right)e^{-x} +x \left ( \frac{\d z}{\d x} -z\right)e^{-x} - ze^{-x} \\ &&&= (x+1) \frac{\d^2 z}{\d x^2} -(x+2)\frac{\d z}{\d x} \\ \Rightarrow && \frac{\d}{\d x} \left ( \frac{\d z}{\d x}\right) &= \left ( \frac{x+2}{x+1}\right) \frac{\d z}{\d x} \end{align*} Therefore \(\frac{\d z}{\d x} = A(x+1)e^x \) and so \begin{align*} z &= A \int (x+1)e^{x} \d x \\ &= A \left ( \left [ (x+1)e^x\right] - \int e^x \d x \right) \\ &= A(x+1)e^x - Ae^x + B \\ y &= Ax + Be^{-x} \end{align*}
3. We have found the complementary solution. To find a particular integral consider \(y = ax^2 + bx + c\), then \(y' = 2ax+b, y'' = 2a\) and we have \begin{align*} && x^2+2x+1 &= 2a(x+1) + x(2ax+b) - (ax^2+bx+c) \\ \Rightarrow && x^2+2x+1 &= ax^2+ 2ax + 2a-c \\ \Rightarrow && a = 1, &c=1 \end{align*} so the general solution should be \[ y = Ax + Be^{-x} + x^2+1 \]

Solution: \begin{align*} && y &= e^x \\ && y' &= e^x \\ && y'' &= e^x \\ \Rightarrow && (x-1)y'' - x y' + y &= (x-1)e^x - xe^x + e^x \\ &&&= 0 \end{align*} Suppose \(y = ue^x\) then \begin{align*} && y' &= u'e^x + ue^x \\ && y'' &= (u''+u')e^x + (u'+u)e^x \\ &&&= (u''+2u' +u)e^x \\ \\ && 0 &= (x-1)y'' - x y' + y \\ &&&= [(x-1)(u''+2u'+u) - x(u'+u)+u]e^x \\ &&&= [(x-1)u'' +(x-2)u']e^x \\ \Rightarrow && 0 &= (x-1)u'' + (x-2)u' \\ v = u': && 0 &= (x-1)v' + (x-2) v \\ \Rightarrow && \frac{v'}{v} &= -\frac{x-2}{x-1} \\ &&&= -1-\frac{1}{x-1} \\ \Rightarrow && \ln v &= -x - \ln(x-1) + C \\ \Rightarrow && v &= A(x-1)e^{-x} \\ && u &= \int Axe^{-x} - Ae^{-x} \d x \\ &&&= \left [-Axe^{-x} +Ae^{-x} \right] + \int Ae^{-x} \d x \\ &&&= -Axe^{-x} + D\\ \Rightarrow && y &= ue^x \\ &&&= -Ax + De^x \end{align*}

Solution: \begin{align*} && \int \frac{{\rm P} ( x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x &= \int \frac{{\rm Q} (x){\rm R}'(x) - {\rm Q}'(x){\rm R}(x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x \\ &&&= \int \frac{\d}{\d x} \left ( \frac{R(x)}{Q(x)} \right) \d x \\ &&&= \frac{R(x)}{Q(x)} + C \end{align*}

1. Suppose \(Q(x) = 1 + 2x + 3x^2, P(x) = 5x^2-4x-3\), and \(R(x) = a +bx + cx^2\), then \begin{align*} && 5x^2-4x-3 &= (1 + 2x + 3x^2)(2cx+b) - (6x+2)(a+bx+cx^2) \\ &&&= (6c-6c)x^3 + (3b+4c-6b-2c)x^2 + \\ &&&\quad+(2c+2b-6a-2b)x + (b-2a) \\ \Rightarrow && 2c-3b &= 5 \\ && 2c-6a &= -4 \\ && b - 2a &= -3 \\ \Rightarrow && b &= 2a - 3\\ && c &= 3a-2 \end{align*} So say \(a = 0, b = -3, c = -2\) we will have \begin{align*} \int \frac{5x^2 - 4x - 3} {(1 + 2x + 3x^2 )^2 } \, \d x &= \frac{-3x-2x^2}{1+2x+3x^2} + C \end{align*} Suppose we have a different value of \(a\), then we end up with: \begin{align*} \frac{a+(2a-3)x+(3a-2)x^2}{1+2x+3x^2} = a +\frac{-3x-2x^2}{1+2x+3x^2} \end{align*} So the different antiderivatives differ by a constant.
2. \(\,\) \begin{align*} && \frac{\d }{\d x} \left ( \frac{1}{1+\cos x + 2 \sin x } y\right) &= \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \\ \end{align*} We want to find \(R(x) = A \cos x + B\sin x + C\) such that \begin{align*} &&5-3\cos x + 4 \sin x &= (1+\cos x + 2 \sin x)R'(x) - R(x)(-\sin x + 2 \cos x) \\ &&&= (1+\cos x + 2 \sin x)(-A\sin x + B \cos x) \\ &&&\quad- (A\cos x + B \sin x + C)(-\sin x + 2 \cos x) \\ &&&=(-A+C) \sin x + (B-2C)\cos x +\\ &&&\quad\quad (2B-A+A-2B)\sin x \cos x \\ &&&\quad\quad (-2A+B)\sin^2x+(B-2A)\cos^2x \\ &&&= (-A+C)\sin x + (B-2C)\cos x +(B-2A) \\ \Rightarrow && B-2A &= 5\\ && C-A &= 4 \\ && B-2C &= -3 \\ \end{align*} There are many solutions so WLOG \(C=4, A = 0, B = 5\) and so \begin{align*} && \int \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \d x &= \frac{5\sin x +4}{1+\cos x + 2 \sin x} + K \\ \Rightarrow && y &= 5\sin x + 4 + K(1 + \cos x + 2 \sin x) \end{align*}

Solution: \begin{align*} && \frac{\d y}{\d x} &= f(x) y + \frac{g(x)}{y} \\ && y \frac{\d y}{\d x} &= f(x) y^2 + g(x) \\ u = y^2: && \frac12 \frac{\d u}{\d x} &= f(x) u + g(x) \end{align*} Which is a linear differential equation for \(u\). \begin{align*} && \frac12 u' &= \frac1x u -1 \\ \Rightarrow && u' - \frac2xu &= -1 \\ \Rightarrow && \frac{1}{x^2} u' - \frac{2}{x^3} u &= -\frac{1}{x^2} \\ \Rightarrow && (\frac{u}{x^2})' &= - \frac{1}{x^2} \\ \Rightarrow && \frac{u}{x^2} &= \frac1x + C \\ \Rightarrow && u &= Cx^2 + x \\ \Rightarrow && y^2 &= Cx^2 + x \end{align*} If \((1,1)\) is on the curve then \(1 = C + 1 \Rightarrow C = 0 \Rightarrow y^2 = x\). If \((2,2)\) is on the curve then \(4 = 4C + 2 \Rightarrow C = \frac12 \Rightarrow y^2 = x + \frac12 x^2\). If \((3,3)\) is on the curve then \(9 = 9C + 3 \Rightarrow C = \frac23 \Rightarrow y^2 = x + \frac23 x^2\)

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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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Solution:

1. \begin{align*} && c &= \left (y+2x \right)^3\left (y-4x \right) \\ \Rightarrow && 0 &= 3\left (y+2x \right)^2\left (y-4x \right)\left ( \frac{\d y}{\d x} + 2 \right) + \left (y+2x \right)^3 \left ( \frac{\d y}{\d x} - 4 \right) \\ \Rightarrow && 0 &= 3(y-4x)\left ( \frac{\d y}{\d x} + 2 \right) + \left (y+2x \right) \left ( \frac{\d y}{\d x} - 4 \right) \\ \Rightarrow &&&= \frac{\d y}{\d x} \left (3(y-4x) + (y+2x) \right) + 6(y-4x)-4(y+2x) \\ &&&= \frac{\d y}{\d x} \left ( 4y-10x\right) + 2y-32x \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{16x-y}{2y-5x} \end{align*}
2. \begin{align*} && c &= \left ( y + ax \right)^n \left ( y + bx \right) \\ \Rightarrow && 0 &= n\left ( y + ax \right)^{n-1} \left ( y + bx \right)\left ( \frac{\d y}{ \d x}+a \right) + \left ( y + ax \right)^{n} \left ( \frac{\d y}{ \d x}+b \right) \\ \Rightarrow && 0 &= n \left ( y + bx \right)\left ( \frac{\d y}{ \d x}+a \right) + \left ( y + ax \right) \left ( \frac{\d y}{ \d x}+b \right) \\ &&&= \frac{\d y}{\d x} \left ( (n+1)y + (nb+a)x \right) + an(y+bx) + by+bax \\ &&&= \frac{\d y}{\d x} \left ( (n+1)y + (nb+a)x \right) + (an+b)y+ab(n+1)x \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{(an+b)y+ab(n+1)x}{(n+1)y+(nb+a)x} \end{align*} We must have \(ab = 10, a+b = -7\) so say \(a=-5,b=-2,n=2\) and we have \((y-5x)^2(y-2) = c\) is our general solution to the differential equation

Solution:

1. \(f(x) = \lfloor x \rfloor+1\)
2. Clearly \(\displaystyle F(x) = \int_0^x f(t) \d t\), in particular: \begin{align*} && F(n) &= \int_0^n f(t) \d t \\ &&&= \sum_{i=1}^n \int_{i-1}^i f(t) \d t \\ &&&= \sum_{i=1}^n \int_{0}^1 f(t-i+1) \d t \\ &&&= \sum_{i=1}^n \int_{0}^1 f(t) \d t \\ &&&= n \int_{0}^1 f(t) \d t\\ &&&= n F(1) \end{align*}
3. \(\,\) \begin{align*} && 0 &= \frac{\d y}{\d x} +f(x) y \\ \Rightarrow && \int -f(x) \d x &= \int \frac1y \d y\\ \Rightarrow && -F(x) & = \ln y + C \\ x=0,y=1: && C &= -F(0) \\ \Rightarrow && y &= \exp(F(0)-F(x)) \end{align*} Well this \(F(0)-F(x)\) is equivalent to \(-F(x)\) where \(F(0) = 0\), in particular \(F(n) = nF(1)\), so \(y(n) = e^{-nF(1)}\) which tends to zero as long as \(F(1) > 0\), but since \(f(x) > 0\) for all \(x\) this must be true.

Solution:

1. Since water flows out a rate proportional to the water in the tank we must have \(\dot{y} = -ky\), ie \(y = Ae^{-k t}\). Since \(t = 0, y = 1\) we have \(y = e^{-kt} = (e^{-k})^t\), so call \(b = e^{-k}\) and we have the result. (Since \(k > 0 \Rightarrow b < 1\)
2. Notice that \begin{align*} && \dot{y} &= -\underbrace{ky}_{\text{flow out}} + \underbrace{a}_{\text{flow in}} \\ &&&= y\ln b + a \\ \Rightarrow && \dot{y} - y \ln b &= a \\ \\ \text{CF}: && y &= Ae^{\ln b t} = Ab^t\\ \text{PI}: && y &= -\frac{a}{\ln b} \\ t = 0, y = 1: && 1 &= A-\frac{a}{\ln b} \\ \Rightarrow && y &= \frac{a}{\ln b} \left ( b^t - 1 \right)+b^t \end{align*}

Solution:

1. \(,\) \begin{align*} u = 1/y, \frac{\d u}{\d x} = -1/y^2 y': && -\frac{\d u}{\d x} + u &= e^{2x} \\ \Rightarrow && \frac{\d}{\d x} \left (e^{-x}u \right) = -e^{x} \\ \Rightarrow && e^{-x}u &= -e^x+C \\ \Rightarrow && u &= Ce^x-e^{2x} \\ \Rightarrow && y &= \frac{1}{Ce^x-e^{2x}} \end{align*}
2. \(\,\) \begin{align*} u = 1/y^2, u' = -2/y^3 y': && -2u'+u &= e^{2x} \\ \Rightarrow && \frac{\d}{\d x} \left (e^{-1/2x}u\right) &= -\frac12e^{\frac32x} \\ \Rightarrow && e^{-\frac12x}u &= -\frac13e^{\frac32x}+C \\ \Rightarrow && u &= -\frac13e^{2x}+Ce^{\frac12x} \\ \Rightarrow && y &= \left ( \frac{1}{Ce^{\frac12x}-\frac13e^{2x}}\right)^{\frac12} \end{align*}