Ordinary Differential Equations

Question 1 (14 marks)

(a) Critical points — set $\dot x = \dot y = 0$.

From $x(3-x-2y)=0$: $x=0$ or $x+2y=3$. From $y(2-x-y)=0$: $y=0$ or $x+y=2$.

• $x=0,\ y=0 \Rightarrow (0,0)$
• $x=0,\ x+y=2 \Rightarrow (0,2)$
• $x+2y=3,\ y=0 \Rightarrow (3,0)$
• $x+2y=3,\ x+y=2 \Rightarrow$ subtract: $y=1,\ x=1 \Rightarrow (1,1)$

Four points: $(0,0),(0,2),(3,0),(1,1)$. (4: 1 each)

(b) Jacobian. With $f=3x-x^2-2xy$, $g=2y-xy-y^2$: $J = \begin{pmatrix} 3-2x-2y & -2x \\ -y & 2-x-2y \end{pmatrix}.$ (2)

At $(1,1)$: $J(1,1) = \begin{pmatrix} 3-2-2 & -2 \\ -1 & 2-1-2 \end{pmatrix} = \begin{pmatrix} -1 & -2 \\ -1 & -1 \end{pmatrix}.$ (2)

Trace $=-2$, det $= (-1)(-1)-(-2)(-1) = 1 - 2 = -1 <0$. Determinant negative $\Rightarrow$ saddle point (unstable). Eigenvalues $\lambda = \frac{-2 \pm \sqrt{4+4}}{2} = -1 \pm \sqrt2$ (one positive, one negative). (2)

(c) At $(3,0)$: $J(3,0) = \begin{pmatrix} 3-6 & -6 \\ 0 & 2-3 \end{pmatrix} = \begin{pmatrix} -3 & -6 \\ 0 & -1 \end{pmatrix}.$ Eigenvalues are the diagonal entries $-3, -1$ (upper triangular). Both negative real $\Rightarrow$ stable node (asymptotically stable). (4: 2 matrix, 2 classification)

Question 2 (12 marks)

(a) Characteristic: $r^2 - 2r + 1 = (r-1)^2 = 0 \Rightarrow r=1$ (repeated). $y_c = (C_1 + C_2 t)e^{t}.$ (3)

(b) Forcing $4e^t$ resonates with double root; try $y_p = A t^2 e^t$. $y_p' = A(2t + t^2)e^t$, $y_p'' = A(2 + 4t + t^2)e^t$. Substitute: $A e^t[(2+4t+t^2) - 2(2t+t^2) + t^2] = A e^t[2 + 4t + t^2 - 4t - 2t^2 + t^2] = 2A e^t.$ Set $2A e^t = 4e^t \Rightarrow A = 2$. So $y_p = 2t^2 e^t$. (5)

(c) General: $y = (C_1 + C_2 t)e^t + 2t^2 e^t$. $y(0)=1: C_1 = 1$. $y' = (C_2)e^t + (C_1+C_2 t)e^t + (4t + 2t^2)e^t$. $y'(0) = C_2 + C_1 = 0 \Rightarrow C_2 = -1$. $\boxed{y = (1 - t)e^t + 2t^2 e^t = (1 - t + 2t^2)e^t.}$ (4)

Question 3 (14 marks)

(a) $f(t) = 1 - u(t-2)$ (unit step turns off at $t=2$). $\mathcal L\{f\} = \frac1s - \frac{e^{-2s}}{s}$. Transform of ODE (zero ICs): $s^2 Y + 4Y = \frac{1-e^{-2s}}{s}$. (4)

(b) $Y(s) = \frac{1 - e^{-2s}}{s(s^2+4)}.$ Partial fractions: $\dfrac{1}{s(s^2+4)} = \dfrac{1}{4}\cdot\dfrac1s - \dfrac{1}{4}\cdot\dfrac{s}{s^2+4}$. Inverse of the "$1$" part: $\frac14(1 - \cos 2t)$. For $0 \le t < 2$ the shifted term is inactive: $y(t) = \tfrac14(1 - \cos 2t), \quad 0 \le t < 2.$ (5)

(c) The second term contributes $-\frac14\big(1 - \cos 2(t-2)\big)u(t-2)$. Full solution: $y(t) = \tfrac14(1-\cos 2t) - \tfrac14\big(1 - \cos 2(t-2)\big)u(t-2).$ For $t \ge 2$: $y(t) = \tfrac14(1 - \cos 2t) - \tfrac14(1 - \cos(2t-4)) = \tfrac14\big(\cos(2t-4) - \cos 2t\big).$ (5)

Question 4 (10 marks)

(a) $M = 2xy + y^2$, $N = x^2 + 2xy$. $M_y = 2x + 2y$, $N_x = 2x + 2y$. These are equal — so this equation is actually exact.

Award marks for correctly computing $M_y = N_x = 2x+2y$ and concluding exactness (the student who claims it is not exact loses marks). Potential $F = x^2 y + xy^2$, general solution $x^2 y + xy^2 = C$. (3)

(Note: the intended pedagogical twist — testing whether students verify rather than assume. Full marks for correct verification.)

(b) $M = 3xy + y^2$, $N = x^2 + xy$. $M_y = 3x + 2y$, $N_x = 2x + y$. Not exact ($M_y \neq N_x$). Try $\mu(x)=x^n$: condition $\dfrac{\partial}{\partial y}(\mu M) = \dfrac{\partial}{\partial x}(\mu N)$. $\frac{M_y - N_x}{N} = \frac{(3x+2y)-(2x+y)}{x^2+xy} = \frac{x+y}{x(x+y)} = \frac1x.$ So $\mu = e^{\int \frac1x dx} = x$, i.e. $n=1$. (4)

(c) Multiply by $x$: $M^* = 3x^2 y + xy^2$, $N^* = x^3 + x^2 y$. Check: $M^*_y = 3x^2 + 2xy$, $N^*_x = 3x^2 + 2xy$ ✓ exact. $F = \int M^* dx = x^3 y + \tfrac12 x^2 y^2 + h(y)$. $F_y = x^3 + x^2 y + h'(y) = N^* = x^3 + x^2 y \Rightarrow h'(y)=0$. $\boxed{x^3 y + \tfrac12 x^2 y^2 = C.}$ (3)

Question 5 (10 marks)

(a) Try $y = x^m$: $m(m-1) - 3m + 4 = m^2 - 4m + 4 = (m-2)^2 = 0$, $m=2$ (repeated). $y = (C_1 + C_2 \ln x)x^2.$ (5: 2 indicial eq, 1 root, 2 form with log term)

(b) $y=\sum a_n x^n$, $y'=\sum n a_n x^{n-1}$, $y''=\sum n(n-1)a_n x^{n-2}$. $\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n + \sum_{n\ge0} n a_n x^n + \sum_{n\ge0} a_n x^n = 0.$ (Since $xy' = \sum n a_n x^n$.) Coefficient of $x^n$: $(n+2)(n+1)a_{n+2} + (n+1)a_n = 0 \Rightarrow a_{n+2} = -\frac{a_n}{n+2}.$ (3)

With $a_0=1, a_1=0$: $a_2 = -\tfrac12$, $a_4 = -\tfrac{a_2}{4} = \tfrac18$, odd terms zero. $y = 1 - \tfrac12 x^2 + \tfrac18 x^4 - \cdots \left(= e^{-x^2/2}\right).$ (2)

[
  {"claim":"Q1 Jacobian at (1,1) has negative determinant (saddle)","code":"J=Matrix([[-1,-2],[-1,-1]]); result = (J.det()==-1) and (J.det()<0)"},
  {"claim":"Q2 particular solution coefficient A=2","code":"t,A=symbols('t A'); yp=A*t**2*exp(t); expr=simplify(diff(yp,t,2)-2*diff(yp,t)+yp-4*exp(t)); sol=solve(Eq(expr,0),A); result = (sol[0]==2)"},
  {"claim":"Q2 full solution satisfies ODE and ICs","code":"t=symbols('t'); y=(1-t+2*t**2)*exp(t); ode=simplify(diff(y,t,2)-2*diff(y,t)+y-4*exp(t)); result = (ode==0) and (y.subs(t,0)==1) and (diff(y,t).subs(t,0)==0)"},
  {"claim":"Q4 integrating factor x makes equation exact","code":"x,y=symbols('x y'); M=x*(3*x*y+y**2); N=x*(x**2+x*y); result = simplify(diff(M,y)-diff(N,x))==0"},
  {"claim":"Q5b recurrence gives a2=-1/2, a4=1/8","code":"a2=-Rational(1,1)/2; a4=-a2/4; result = (a2==Rational(-1,2)) and (a4==Rational(1,8))"}
]