Problems

---

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:

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*} && g(0) &= f(0)+f(-0) = 2f(0) = 2 \\ && g'(x) &= f'(x) - f'(-x) \\ && g'(0) &= f'(0) - f'(-0) = 0 \\ && g''(x) &= f''(x) +f''(-x) \\ \Rightarrow && g''(x) + g(x) &= f''(x) +f''(-x) + f(x) + f(-x) \\ &&&= f''(x)+ f(-x) +f''(-x) + f(x) \\ &&&= x + 3 \cos 2x + (-x + 3 \cos (-2x) ) \\ &&&= 6 \cos 2x \\ \end{align*} Considering the homogeneous part, we should expected a solution of the form \(g(x) = A \sin x + B \cos x\). Seeking an integrating factor of the form \(g(x) = C \cos 2x\) we see that \(-4C \cos 2x + C \cos 2x = 6 \cos 2x \Rightarrow -3C = 6 \Rightarrow C = -2\). Therefore the general solution is \begin{align*} && g(x) &= A\sin x + B \cos x - 2\cos 2x \\ && g(0) &= B - 2 = 2\\ && g'(0) &= A = 0 \\ \Rightarrow && g(x) &= 4\cos x - 2\cos 2x \\ \end{align*} \begin{align*} && h(0) &= f(0) - f(-0) = 0 \\ && h'(x) &= f'(x) + f'(-x) \\ && h'(0) &= f'(0) + f'(-0) = -2 \\ && h''(x) &= f''(x) - f''(-x) \\ \Rightarrow && h''(x) - h(x) &= f''(x) - f''(-x) -( f(x) - f(-x)) \\ &&&= f''(x) +f(-x)- (f''(-x) + f(x)) \\ &&&= x + 3\cos 2x - (-x + 3 \cos(-2x)) \\ &&&= 2x \end{align*} Considering the homogeneous part, we should expect a solution of the form \(Ae^x + Be^{-x}\). For a specific integral, we can take \(-2x\), ie \begin{align*} && h(x) &= Ae^x + Be^{-x} - 2x \\ && h(0) &= A+B =0 \\ && h'(0) &= A-B-2 =-2 \\ \Rightarrow && A &=B = 0 \\ \Rightarrow && h(x) &= -2x \end{align*} Therefore \(f(x) = \frac12(f(x) + f(-x)) + \frac12(f(x) -f(-x)) = 2\cos x - \cos 2x -x\)

---

Solution: The total volume in the first vat at time \(t\) is always \(K\), since \(b\) litres per second are coming in and \(b\) litres per second are going out. \begin{align*} &&\frac{\d V}{\d t} &= \underbrace{a}_{\text{incoming chemical }A} - \underbrace{b}_{\text{outgoing volume}} \cdot \underbrace{\frac{V(t)}{K}}_{\text{fraction of outgoing which is }A} \\ &&&= a - b \frac{V}{K} \\ \Rightarrow && \int \frac{1}{a-b\frac{V}{K}}\d V &= \int \d t \\ && - \frac{K}{b} \ln |a - b \frac{V}{K}| &= t +C\\ (t,V) = (0,0): && -\frac{K}{b} \ln a &= C \\ \Rightarrow && 1-\frac{b}{a} \frac{V}{K} &= e^{-bt/K} \\ \Rightarrow && V &= \frac{aK}{b} (1 - e^{-bt/K}) \end{align*} \begin{align*} &&\frac{\d W}{\d t} &= \underbrace{b}_{\text{incoming volume}} \cdot \underbrace{\frac{a}{b} (1 - e^{-bt/K})}_{\text{incoming fraction }A} - \underbrace{b}_{\text{outgoing volume}} \cdot \underbrace{ \frac{W(t)}{L}}_{\text{fraction of outgoing which is }A} \\ &&&= a (1 - e^{-bt/K}) - b \frac{W}{L} \\ \Rightarrow && \frac{\d W}{\d t} + \frac{b}{L} W &= a (1-e^{-bt/K}) \\ && \frac{\d}{\d t} \left ( e^{b/L t} W\right) &= ae^{b/L t}(1-e^{-bt/K}) \\ \Rightarrow && W &= e^{-bt/L} \left ( \frac{aL}{b}e^{b/Lt} - \frac{a}{\frac{b}{L} - \frac{b}{K}}e^{b/L t - b/K} \right) + Ce^{-bt/L} \\ &&&= \frac{aL}{b} \left (1 - \frac{K}{K-L}e^{-b/Kt} \right)+ Ce^{-bt/L} \\ (t,W) = (0,0): && 0 &= \frac{aL}{b} \frac{-L}{K-L} + C \\ \Rightarrow && C &= \frac{aL^2}{b(K-L)} \\ \Rightarrow && W &= \frac{aL}{b} \left (1 - \frac{K}{K-L} e^{-bt/K} + \frac{L}{K-L} e^{-bt/L} \right) \end{align*}

---

Solution: First we seek the complementary function. \begin{align*} && \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 0 \\ \Rightarrow && r &= A \sin 2\theta + B \cos 2 \theta \end{align*} Next we seek a particular integral, of the form \(r = C \sin 3 \theta\). \begin{align*} && \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 5 \sin 3 \theta \\ \Rightarrow && -9C \sin 3 \theta + 4C \sin 3 \theta &= 5 \sin 3 \theta \\ \Rightarrow && C &= -1 \\ \end{align*} So our general solution is \(A \sin 2\theta + B \cos 2 \theta -\sin 3 \theta\). Plugging in boundary conditions we obtain: \begin{align*} \theta = \frac{\pi}{2}, r = 1: &&1 &= -B +1 \\ \Rightarrow && B &= 0 \\ \theta = \frac{\pi}{2}, \frac{\d r}{\d \theta} = -2: && -2 &= -2A \\ \Rightarrow && A &= 1 \end{align*} So the general solution is \(r = \sin 2 \theta - \sin 3 \theta = 2 \sin \left ( \frac{-\theta}{2} \right) \cos \left (\frac{5 \theta}{2} \right)\) First notice that for \(\theta \in \left [\frac{\pi}{5}, \frac{3 \pi}{5} \right]\) this is positive, and it is zero on the end points, therefore we are tracing out a a loop. The area of the loop will be: \begin{align*} A &= \int_{\pi/5}^{3\pi/5} \frac12 \left ( \sin 2 \theta - \sin 3 \theta \right)^2 \d \theta \\ &= \frac12\int_{\pi/5}^{3\pi/5} \sin^2 2\theta + \sin^2 3 \theta - 2 \sin 2 \theta \cos 3 \theta \d \theta \\ &= \frac12\int_{\pi/5}^{3\pi/5} \frac{1-2 \cos 4 \theta}{2} + \frac{1-2 \cos6 \theta}{2} - \sin5 \theta-\cos\theta \d \theta \\ &= \frac12 \left [\theta - \frac14 \sin 4 \theta-\frac16 \sin 6 \theta + \frac15 \cos 5 \theta - \sin \theta \right]_{\pi/5}^{3\pi/5} \\ &= \frac{\pi}{5} +\frac{25}{48}\left [ \sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right) \right] \end{align*}

---

Solution: \begin{align*} && \frac{\d x}{\d t} &= 4(x-y) \\ && \frac{\d y}{\d t} &= x - 12(e^{2t}+e^{-2t}) \\ \Rightarrow && \frac{\d^2 x}{\d t^2} &= 4 \frac{\d x}{\d t}-4\frac{\d y}{\d t} \\ &&&= 4 \frac{\d x}{\d t}-4 \left ( x - 12(e^{2t}+e^{-2t}) \right) \\ \Rightarrow && \frac{\d^2 x}{\d t^2} - 4 \frac{\d x}{\d t}+4x &= 48 (e^{2t}+e^{-2t}) \end{align*} This differential equation has characteristic polynomial \(\lambda^2 - 4\lambda + 4 = (\lambda-2)^2\). Therefore we should expect a general solution of \((At+B)e^{2t}\). For particular integrals we should try \(ke^{-2t}\) and \(Ct^2 e^{2t}\). For the former, we have: \begin{align*} && 48 &= 4k+8k+k \\ \Rightarrow && k &= \frac{48}{13} \end{align*} For the latter we have: \begin{align*} &&4Ct^2e^{2t} -4C(2te^{2t}+2t^2e^{2t})+2C((1+2t)e^{2t}+2t^2e^{2t}) &= 48e^{2t} \\ \Rightarrow && 2C &= 48 \\ \Rightarrow && C &= 24 \end{align*} Therefore the solution should be: \begin{align*} x = (At+B)e^{2t} + \frac{48}{13}e^{-2t} + 24t^2 e^{2t} \\ x(0) = B + \frac{48}{13} \\ x'(0) = 2B+A-\frac{96}{13} \\ x =\frac{48}{13}((4t-1)e^{2t}+e^{-2t})+24t^2e^{2t} \\ y = x - \frac{1}{4} \frac{\d x}{\d t} \end{align*}

---

Solution:

1. \(M(0) = 0\) since if there's no space on the shelf, we wont be able to put any books on the shelf.
2. If the shelf has length \(1\) it can only fit a thin book. For a thin book to be placed on the shelf, the very first book which comes to be placed must be thin. But this happens with probability \(q\). Therefore \(M(1) = q\).
3. Suppose no books have been placed on the shelf, then with probability \(p\) a large book gets placed on the shelf, and the expected number of books to be placed on the shelf is equivalent to how many books will be placed on the shelf if the shelf only had \(n-2\) spaces. This is \(M(n-2)\). Similar if the book which arrives first is thin (with probability \(q\)) then there will be \(M(n-1)\) more books placed on the shelf in expectation. We've just added \(1\) more book, therefore \(M(n) = 1+pM(n-2) + qM(n-1)\) or rearranging \(M(n) - qM(n-1) - pM(n-2) = 1\).

Suppose \(M(n) = (-p)^n\), notice that: \begin{align*} M(n) - qM(n-1) - pM(n-2) &= (-p)^n - (1-p)(-p)^n - p(-p)^{n-2} \\ &= (-p)^{n-2}(p^2+(1-p)p-p) \\ &= 0 \end{align*} Suppose \(M(n) = B\), notice that: \begin{align*} M(n) - qM(n-1) - pM(n-2) &= B - (1-p)B - pB \\ &= 0 \end{align*} Finally, if \(M(n) = Cn\) notice that: \begin{align*} M(n) - qM(n-1) - pM(n-2) &= Cn - (1-p)C(n-1) - pC(n-2) \\ &= C(n(1-(1-p)+p)+(1-p)+2p) \\ &= C(1+p) \end{align*} Therefore if \(C = \frac{1}{1+p}\) we have that: \(M(n) = A(-p)^n + B + Cn\) satisfies our recurrence. We also need \(M(0) = 0\) and \(M(1) = q\) \begin{align*} 0 &= M(0) \\ &= A + B \\ 1-p &= M(1) \\ &= -pA+B \end{align*} \((1+p)A = p-1 \Rightarrow A = \frac{p-1}{1+p}, B = \frac{1-p}{1+p}\). Therefore: \[ M(n) = -\frac{1-p}{1+p}(-p)^n + \frac{1-p}{1+p} + \frac{n}{1+p} \] is a possible solution to this equation

---

Solution: Letting \(\mathbf{x} =\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}\) and \(\mathbf{A} = \begin{pmatrix} -2 & 5 \\ -a & 2 \end{pmatrix}\) then our differential equation is \(\mathbf{x}' = \mathbf{Ax} + \begin{pmatrix} 0 \\2 \cos t \end{pmatrix}\). Looking at the eigenvalues of \(\mathbf{A}\), we find: \begin{align*} && \det \begin{pmatrix} -2-\lambda & 5 \\ -a & 2 -\lambda \end{pmatrix} &= (\lambda^2-4)+5a\\ &&&= \lambda^2 +5a-4 \end{align*} Therefore if \(a = 1\), \(\lambda = \pm i\). In which case we should expect the complementary solutions to be of the form \(\mathbf{x} = \begin{pmatrix} A \sin t + B \cos t \\ C \sin t + D \cos t \end{pmatrix}\). The first equation tells us that \((A-5D+B)\cos t + (-B+5C)\sin t=0\) so the complementary solution is:\(\mathbf{x} = \begin{pmatrix} 5(D-C) \sin t + 5C \cos t \\ C \sin t + D \cos t \end{pmatrix}\). Looking for a particular integral, we should expect to try something like \(\mathbf{x} = \begin{pmatrix} Et\cos t+Ft\sin t\\ Gt\cos t+Ht \sin t\end{pmatrix}\) and we find