Ordinary Differential Equations

Question 1

(a) Multiply the ODE by an unknown $\mu(x)$: $\mu y' + \mu P y = \mu Q.$ We want the LHS to equal $\frac{d}{dx}(\mu y) = \mu y' + \mu' y$. (1) Matching the $y$ terms requires $\mu' = \mu P. \quad\text{(1 — the key condition)}$ This is separable: $\frac{d\mu}{\mu} = P\,dx \Rightarrow \ln\mu = \int P\,dx$, (1) $\boxed{\mu(x) = e^{\int P(x)\,dx}}. \quad (1)$ Then $(\mu y)' = \mu Q \Rightarrow y = \frac{1}{\mu}\int \mu Q\,dx$. (1)

(b) Here $P=2/x$, so $\int P\,dx = 2\ln x$, $\mu = x^2$. (1) $(x^2 y)' = x^2\cdot\frac{\cos x}{x^2} = \cos x. \quad (1)$ Integrate: $x^2 y = \sin x + C$. (1) Apply $y(\pi)=0$: $\pi^2\cdot 0 = \sin\pi + C = 0 + C \Rightarrow C=0$. (1) $\boxed{y = \frac{\sin x}{x^2}}. \quad (1)$

Question 2

(a) For $ay''+by'+cy=0$ with constant coefficients, derivatives of $e^{rx}$ reproduce $e^{rx}$ times a constant, so the equation collapses to an algebraic one: $a r^2 e^{rx} + b r e^{rx} + c e^{rx} = e^{rx}(ar^2+br+c)=0.$ Since $e^{rx}\ne0$, we need $ar^2+br+c=0$ — the characteristic equation. (2)

Repeated root case: if $r$ is a double root, $r=-b/(2a)$ and we only get one solution $y_1=e^{rx}$. Seek $y_2 = v(x)e^{rx}$ (reduction of order). (1) Then $y_2'=(v'+rv)e^{rx}$, $y_2''=(v''+2rv'+r^2v)e^{rx}$. Substitute: $a(v''+2rv'+r^2v)+b(v'+rv)+cv=0$ $a v'' + (2ar+b)v' + (ar^2+br+c)v = 0.$ (1) The $v$-coefficient is $0$ (root), and $2ar+b=0$ (double root $\Rightarrow r=-b/2a$). (1) So $a v''=0 \Rightarrow v''=0 \Rightarrow v = C_1 + C_2 x$; the new piece is $v=x$, giving $y_2 = x e^{rx}$. (1)

(b) Characteristic: $r^2-4r+13=0 \Rightarrow r = \frac{4\pm\sqrt{16-52}}{2} = 2\pm 3i$. (2) Complex solutions: $e^{(2+3i)x}, e^{(2-3i)x}$. Using Euler $e^{i\theta}=\cos\theta+i\sin\theta$: (1) $e^{2x}e^{\pm 3ix} = e^{2x}(\cos 3x \pm i\sin 3x).$ Real combinations (sum/difference over 2, 2i) give $e^{2x}\cos3x$ and $e^{2x}\sin3x$. (1) $\boxed{y = e^{2x}(C_1\cos 3x + C_2\sin 3x)}. \quad (2)$

Question 3

(a) $\mathcal{L}\{e^{at}\} = \int_0^\infty e^{-st}e^{at}\,dt = \int_0^\infty e^{-(s-a)t}\,dt.$ (1) $= \left[\frac{-1}{s-a}e^{-(s-a)t}\right]_0^\infty = 0 - \left(\frac{-1}{s-a}\right) = \frac{1}{s-a}. \quad (1)$ The limit at $\infty$ is $0$ only if $s-a>0$, so ROC is $s>a$. (1)

(b) $\mathcal{L}\{f'\}=\int_0^\infty e^{-st}f'(t)\,dt.$ Integrate by parts, $u=e^{-st}, dv=f'dt$: (1) $=\Big[e^{-st}f(t)\Big]_0^\infty + s\int_0^\infty e^{-st}f(t)\,dt.$ (1) Boundary term: at $\infty$, $\to 0$ (for $s$ in ROC); at $0$, $-f(0)$. (1) $= -f(0) + sF(s) = sF(s)-f(0). \quad (1)$

(c) Transform $y'-3y=0$: $sY-y(0)-3Y=0 \Rightarrow (s-3)Y = 2$. (1) $Y = \frac{2}{s-3}. \quad (1)$ Invert (part a with $a=3$): $y=2e^{3t}$. (1)

Question 4

(a) Transform: $s^2Y - sy(0)-y'(0) + Y = \mathcal{L}\{u_\pi\} = \dfrac{e^{-\pi s}}{s}$. (2) With zero initial conditions: $(s^2+1)Y = \frac{e^{-\pi s}}{s} \Rightarrow Y = \frac{e^{-\pi s}}{s(s^2+1)}. \quad (2)$

(b) Partial fractions: $\dfrac{1}{s(s^2+1)} = \dfrac{1}{s} - \dfrac{s}{s^2+1}$. (2) Inverse of this factor: $1 - \cos t$. (1) The $e^{-\pi s}$ triggers the second shift theorem: multiply by $u_\pi(t)$ and replace $t\to t-\pi$: (1) $y(t) = u_\pi(t)\big[1 - \cos(t-\pi)\big]. \quad (1)$ Piecewise: $\boxed{y(t) = \begin{cases} 0, & 0\le t<\pi \\ 1 - \cos(t-\pi) = 1+\cos t, & t\ge\pi \end{cases}} \quad (1)$

(c) Since $\cos(t-\pi) = -\cos t$, for $t>\pi$: $y(t) = 1 + \cos t. \quad (2)$

Question 5

(a) $\det(A-\lambda I) = (1-\lambda)^2 - 4 = 0 \Rightarrow (1-\lambda)=\pm2 \Rightarrow \lambda_1=3,\ \lambda_2=-1$. (2) For $\lambda=3$: $(A-3I)=\begin{pmatrix}-2&2\\2&-2\end{pmatrix}\Rightarrow v_1=(1,1)^T$. (1) For $\lambda=-1$: $\begin{pmatrix}2&2\\2&2\end{pmatrix}\Rightarrow v_2=(1,-1)^T$. (1) $\mathbf{x}(t)=C_1 e^{3t}\begin{pmatrix}1\\1\end{pmatrix} + C_2 e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}. \quad (1)$

(b) Eigenvalues real, opposite signs ($3>0$, $-1<0$) ⇒ saddle point, which is unstable. (3)

(c) For $\mathbf{x}'=A\mathbf{x}$, let $T=\operatorname{tr}A$, $D=\det A$, discriminant $\Delta=T^2-4D$.

• $D<0$: saddle (unstable).
• $D>0,\ \Delta>0$: node (stable if $T<0$, unstable if $T>0$).
• $D>0,\ \Delta<0$: spiral (stable if $T<0$, unstable if $T>0$); $T=0$ ⇒ centre. (2)

Question 6

(a)

def euler(f, x0, y0, h, n):
    xs, ys = [x0], [y0]
    x, y = x0, y0
    for _ in range(n):
        y = y + h * f(x, y)
        x = x + h
        xs.append(x)
        ys.append(y)
    return xs, ys

Marks: correct update rule $y\leftarrow y+h f(x,y)$ (2), loop/n steps (1), returns both lists (1).

(b) $f(x,y)=y$, $h=0.5$:

• Step 1: $y_1 = 1 + 0.5(1) = 1.5$.
• Step 2: $y_2 = 1.5 + 0.5(1.5) = 2.25$.

Euler estimate $y(1)\approx 2.25$; exact $e\approx 2.71828$. Underestimate (error $\approx0.468$). (2)

[
  {"claim":"Q1b solution y=sin x /x^2 satisfies ODE and IC",
   "code":"x=symbols('x'); y=sin(x)/x**2; lhs=diff(y,x)+2/x*y; result=(simplify(lhs-cos(x)/x**2)==0) and (y.subs(x,pi)==0)"},
  {"claim":"Q2b roots of char eqn are 2+-3i",
   "code":"r=symbols('r'); sols=solve(r**2-4*r+13,r); result=set(sols)==set([2+3*I,2-3*I])"},
  {"claim":"Q4 partial fraction 1/(s(s^2+1))=1/s - s/(s^2+1)",
   "code":"s=symbols('s'); result=simplify(1/(s*(s**2+1)) - (1/s - s/(s**2+1)))==0"},
  {"claim":"Q5 eigenvalues of A are 3 and -1",
   "code":"A=Matrix([[1,2],[2,1]]); result=sorted(A.eigenvals().keys())==[-1,3]"},
  {"claim":"Q6 two-step Euler for y'=y gives 2.25",
   "code":"y=1.0; h=0.5\nfor _ in range(2): y=y+h*y\nresult=abs(y-2.25)<1e-9"}
]