1. We have that $[d] = MLT^{-2}$, $[r]=L$, $[v]=LT^{-1}$, $[\rho] =ML^{-3}$, and $[\mu] = ML^{-1}T^{-1}$. The ansatz: $\begin{align*} MLT^{-2} = [d] &= [r^w v^x \rho^y \mu^z] = [r]^w[v]^x[\rho]^y[\mu]^z = L^wL^xT^{-x}M^yL^{-3y}M^{z}L^{-z}T^{-z}\\&=M^{y+z}L^{w+x-3y-z}T^{-x-z} \end{align*}$for some $w,x,y,z \in \mathbb{Z}$ yields the following system of linear equations:$\begin{align} &L: &&1 = w+x-3y-z\\ &T: &&2 = x+z \\ &M: &&1 = y + z \end{align}$From second & third equations $x=2-z$ and $y = 1-z$. Substituting for $x, y$ in the first equation yields$\begin{align}1 &= w+2-z-3+3z-z\\ &= w+z-1\\ \text{So }2 &=w+z=x+z \text{ (using third equation)} \\ \implies w&=x \end{align}$So $w = x = 2-z$ and $y=1-z$. Hence $[d]=[r]^{2-z}[v]^{2-z}[\rho]^{1-z}[\mu]^{z}= \left[r^{2} v^{2} \rho \left(\frac{\mu}{rv\rho}\right)^{z}\right]$Thus the reduced problem is $d = r^{2} v^{2} \rho \, \tilde{u} ( \frac{\mu}{rv\rho}).$ 2. `(a)` $\frac{d}{dt}x(t) = \frac{\beta}{x(t)}$ `(b)` We have that $\frac{d}{d t} x(t)=\frac{\beta}{x(t)} \text { and } x(0)=1,\, x(4)=3$ $ \begin{align} \text { So } \int_1^{x(t)} \tilde{x} \, d \tilde{x}&=\int_0^t \beta \,d \tilde{t} \\ \left[\frac{1}{2} \tilde{x}^{2}\right]_{1}^{x(t)} &= [\beta \tilde{t}]_{0}^{t} \\ \frac{1}{2} x(t)^2-\frac{1}{2}&=\beta t \\ \text { So } x(t) &=\sqrt{2 \beta t+1} \\ \text{Now } 3 &= x(4)=\sqrt{2\times\beta\times4+1} \\ \implies 9 &= 8\beta+1 \implies \beta = 1.\\ \\ \text{So we have } x(t) &= \sqrt{2t+1} \\ \text{and } x(12) &= \sqrt{2\times12+1}=\sqrt{25}=5 \end{align}$ Therefore the ice will be 5mm thick at 12 noon. `(c)` $\begin{align} \text{Now we have }\frac{d}{dt}x\left(t\right)&=\frac{\beta\left(1+\frac{1}{2}\sin\left(wt\right)\right)}{x(t)} \text{ and }x(0)=1,\,x\left(4\right)=3\\ \text{So }\int_{1}^{x(t)}\tilde{x}\text{~d}\tilde{x}&=\int_{0}^{t}\beta\left(1+\frac{1}{2}\sin(w\tilde{t})\right)d\tilde{t}\\ \left[\frac{1}{2}\tilde{x}^{2}\right]_{1}^{x(t)}&=\beta\left[\tilde{t}-\frac{1}{2w}\cos(w\tilde{t})\right]_{0}^{t} \\ \frac{1}{2}x(t)^{2} - \frac{1}{2}&=\beta \left(t - \frac{1}{2\omega} cos(\omega t) + \frac{1}{2\omega} \right) \\ x(t)^{2} &= \beta \left(2t - \frac{cos(\omega t)}{\omega} + \frac{1}{\omega} \right)+1 \\ \text{So } \quad x(t)&= \sqrt{\beta \left(2t - \frac{cos(\omega t)-1}{\omega} \right)+1} \\ \text{Now } 3&=x(4)=\sqrt{\beta\left(8- \frac{24(cos(\frac{\pi}{3})-1)}{2\pi} \right)+1} \\ \\ \implies 9 &= \left(8+ \frac{6}{\pi}\right) \beta +1 \implies \beta = \frac{8\pi}{6+8\pi} \\ \text{Hence }\quad x(t)&= \sqrt{\frac{8\pi}{6+8\pi} \left(2t - \frac{cos(\omega t)-1}{\omega} \right)+1} \\ \end{align}$ 3. The particular integral is given by: $y_{p}(t)=Asin(\theta t) + Bcos(\theta t)$ with some $A,B \in \mathbb{R}$. Now $\begin{align*} \frac{d}{dt}y_{p}(t) &= A\theta cos (\theta t) - B\theta sin(\theta t)\\ \text{and } \frac{d^{2}}{dt^{2}}y_{p}(t) &= -A\theta^{2}sin(\theta t) - B\theta^{2}cos(\theta t) \\ \end{align*}$Substituting for $y_{p}(t)$ and its derivatives in the DE yields:$\begin{align*} -\theta^{2} (Asin(\theta t) + Bcos(\theta t)) + \eta \theta (Acos(\theta t) - Bsin(\theta t))+b(Asin(\theta t) + Bcos(\theta t)) &= fcos(\theta t)\\ (Ab - B\eta \theta - A\theta^{2})sin(\theta t) + (Bb - A\eta \theta - B\theta^{2})cos(\theta t) &= fcos(\theta t) \end{align*}$By comparing coefficients:$\begin{align*} Ab - B\eta \theta - A\theta^{2} &= 0 \implies A(b - \theta^{2}) = B \eta \theta \implies B = \frac{b-\theta^{2}}{\eta \theta} A\\ Bb - A\eta \theta - B\theta^{2} &= f \implies B(b-\theta^{2}) - A\eta \theta = f\\ \end{align*}$Substituting for B in the second equation gives:$\begin{align*} \frac{(b-\theta^{2})^{2}}{\eta \theta} A -A\eta \theta = f \implies A \left( \frac{(b-\theta^{2})^{2}-\eta^{2} \theta^{2}}{\eta \theta}\right) = f\\ \text{So } A = \frac{\eta \theta f}{(b-\theta^{2})^{2} - \eta^{2} \theta^{2}} \text{ and } B = \frac{(b-\theta^{2}) f}{(b-\theta^{2})^{2} - \eta^{2} \theta^{2}} \end{align*}$ 4. (a) ![[Pasted image 20231115160533.png]] (b)The solution is given by n = 6. As evidence, the graphs below are n = 6 and n = 5 respectively. ![[Screenshot 2023-11-15 at 16.14.32.png]]![[Screenshot 2023-11-15 at 16.16.05.png]]