Ordinary Differential Equations

Q1. (3 marks) (a) Highest derivative $y''$ → order 2. Highest power of $y''$ is 3 → degree 3. Contains $y\,y'$ (product of $y$ and derivative) → nonlinear. (1.5) (b) Order 1, degree 1, linear — coefficients depend only on $x$, $y$ appears to first power. (1.5)

Q2. (4 marks) Separate: $y\,dy = x\,dx$. (1) Integrate: $\tfrac{y^2}{2} = \tfrac{x^2}{2} + C$ → $y^2 = x^2 + C_1$. (1) Apply $y(0)=2$: $4 = 0 + C_1 \Rightarrow C_1 = 4$. (1) Explicit: $y = \sqrt{x^2 + 4}$ (positive root since $y(0)=2>0$). (1)

Q3. (5 marks) Integrating factor $\mu = e^{\int 2\,dx} = e^{2x}$. (1) Multiply: $(e^{2x}y)' = e^{2x}e^{-x} = e^{x}$. (2) Integrate: $e^{2x}y = e^{x} + C$. (1) Solution: $y = e^{-x} + Ce^{-2x}$. (1)

Q4. (5 marks) Characteristic equation: $r^2 - 5r + 6 = 0$. (2) Factor: $(r-2)(r-3)=0 \Rightarrow r = 2, 3$. (2) General solution: $y = C_1 e^{2x} + C_2 e^{3x}$. (1)

Q5. (4 marks) Characteristic: $r^2 + 4 = 0 \Rightarrow r = \pm 2i$. (2) Complex roots $\alpha \pm i\beta$ with $\alpha=0,\beta=2$. (1) $y = C_1\cos 2x + C_2\sin 2x$. (1)

Q6. (4 marks) $\partial M/\partial y = 2x$, $\partial N/\partial x = 2x$ → equal, so exact. (1) $F = \int M\,dx = x^2 y + g(y)$. (1) $\partial F/\partial y = x^2 + g'(y) = N = x^2 + 3y^2 \Rightarrow g'(y)=3y^2 \Rightarrow g=y^3$. (1) Implicit solution: $x^2 y + y^3 = C$. (1)

Q7. (4 marks) $\mathcal{L}\{e^{3t}\} = \dfrac{1}{s-3}$, ROC $s>3$. (2) $\mathcal{L}\{\sin 2t\} = \dfrac{2}{s^2+4}$, ROC $s>0$. (2)

Q8. (4 marks) Complete square: $s^2+4s+13 = (s+2)^2 + 9$. (1) Rewrite numerator: $s+1 = (s+2) - 1$. (1) $\dfrac{s+2}{(s+2)^2+9} - \dfrac{1}{3}\cdot\dfrac{3}{(s+2)^2+9}$. (1) Inverse: $e^{-2t}\cos 3t - \tfrac{1}{3}e^{-2t}\sin 3t$. (1)

Q9. (4 marks) If $f(x,y)$ is continuous on a rectangle $R$ containing $(x_0,y_0)$ (1), and satisfies a Lipschitz condition in $y$ (equivalently $\partial f/\partial y$ continuous) on $R$ (2), then there exists a unique solution $y(x)$ to the IVP on some interval $|x-x_0|<h$ (1).

Q10. (3 marks) Bernoulli with $n=2$; substitution $v = y^{1-n} = y^{-1}$. (1) Then $v' = -y^{-2}y'$; dividing original by $y^2$ and substituting gives (1): $\dfrac{dv}{dx} - v = -1$. (1)

[
  {"claim":"Q2 IVP solution y=sqrt(x^2+4) satisfies ODE and IC","code":"x=symbols('x'); y=sqrt(x**2+4); ode=simplify(diff(y,x)-x/y); ic=y.subs(x,0)-2; result=(ode==0) and (ic==0)"},
  {"claim":"Q3 general solution satisfies y'+2y=e^-x","code":"x,C=symbols('x C'); y=exp(-x)+C*exp(-2*x); result=simplify(diff(y,x)+2*y-exp(-x))==0"},
  {"claim":"Q4 roots of r^2-5r+6 are 2 and 3","code":"r=symbols('r'); result=set(solve(r**2-5*r+6,r))=={2,3}"},
  {"claim":"Q8 inverse Laplace gives e^-2t(cos3t - sin3t/3)","code":"s,t=symbols('s t',positive=True); F=(s+1)/(s**2+4*s+13); f=exp(-2*t)*(cos(3*t)-sin(3*t)/3); result=simplify(laplace_transform(f,t,s,noconds=True)-F)==0"}
]