2019 Paper Ⅲ
PDF
  1. (a) [6 marks] For $x>0$, solve the initial value problem$$x \ln y \frac{\mathrm{d} y}{\mathrm{d} x}=y \ln x, \quad y(1)=2,$$giving $y(x)$ explicitly in terms of $x$.
    (b) [6 marks] Solve the initial value problem$$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+9 y=2 \cos 3 x, \quad y(0)=1, y'(0)=0 .$$(c) [8 marks] Consider the nonlinear differential equation$$\frac{\mathrm{d} y}{\mathrm{d} x}+P(x) y=Q(x) y^{n},$$where $n ≠ 0,1$ and $P(x)$ and $Q(x)$ are suitable arbitrary functions of $x$.
    (i) Show that the substitution $v=y^{1-n}$ transforms the nonlinear differential equation into one which is linear in $v$.
    (ii) Hence find the general solution of$$\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{y}{x}=x^{5} y^{4},$$giving $y(x)$ explicitly in terms of $x$.
    Solution.
  2. (a) [7 marks] If $x=u \cos α-v \sin α$ and $y=u \sin α+v \cos α$, where $α$ is a constant, show that$$\frac{\partial^{2} F}{\partial x^{2}}+\frac{\partial^{2} F}{\partial y^{2}}=\frac{\partial^{2} F}{\partial u^{2}}+\frac{\partial^{2} F}{\partial v^{2}},$$where $F$ is a twice continuously differentiable but otherwise arbitrary function of two variables.
    (b) [6 marks] Find the directional derivative of $F=x y^{2} z^{3}$ along the curve$$
    x=2 \cos t+1, \quad y=3 \sin t-1, \quad z=e^{-t}
    $$at the point $P$ where $t=0$.
    What does the sign of your answer tell you about $F$ at $P$?
    (c) [7 marks] Use the change of variables $ϕ=x+y, ψ=x y$ to show that if $u(x, y)$ is twice continuously differentiable and satisfies the partial differential equation$$x \frac{\partial^{2} u}{\partial x^{2}}-(x+y) \frac{\partial^{2} u}{\partial x \partial y}+y \frac{\partial^{2} u}{\partial y^{2}}+\frac{(x+y)}{(y-x)}\left(\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}\right)=0,$$where $y ≠ x$, then $u(ϕ, ψ)$ satisfies the partial differential equation$$\frac{\partial^{2} u}{\partial ϕ \partial ψ}=0$$Hence find the general solution $u(x, y)$.
    Solution.
  3. (a) [8 marks] (i) State the second-order Taylor expansion of a function $f(x, y)$ around a critical point $\left(x_{0}, y_{0}\right)$.
    (ii) Assuming $f_{x x} ≠ 0$, show that the expansion can be written in the form$$f(x, y)-f\left(x_{0}, y_{0}\right)=\frac{1}{2}\left[f_{x x}\left(\left(x-x_{0}\right)+D_{1}\left(y-y_{0}\right)\right)^{2}+D_{2}\left(y-y_{0}\right)^{2}\right]$$where $f_{x x}, f_{x y}$ and $f_{y y}$ are the second-order partial derivatives of $f$ evaluated at the critical point, and the constants $D_1$ and $D_2$ should be given in terms of these derivatives.
    (iii) Hence explain how the signs of $f_{x x}$ and $f_{x x} f_{y y}-f_{x y}^{2}$ classify the critical point. [You need not consider the case $D_{2}=0$.]
    (b) [5 marks] Find and classify the critical points of the function$$f(x, y)=x^3+2y^3-6x-18 y+5$$(c) Let $\mathbf{F}=\left(x^{2}+y z\right) \mathbf{i}+\left(x y-z^{2}\right) \mathbf{j}+(x-3 z) \mathbf{k} .$ Evaluate the line integral $I=\int_{C} \mathbf{F} · \mathrm{d} \mathbf{r}$ where $C$ is the path given by the straight line from $(0,0,0)$ to $(0,0,1)$ followed by the straight line from $(0,0,1)$ to $(0,1,1)$ followed by the straight line from $(0,1,1)$ to $(1,1,1)$.
    Solution.
  4. Solution.
  5. Solution.
  6. Solution.