Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 6

The characteristic polynomial of this constant-coefficient differential equation is $r^2-r-2=(r-2)(r+1)$. Its distinct roots give

For the forcing $8e^{-2x}$, the method of undetermined coefficients gives the particular term $2e^{-2x}$. The term $e^{-x}$ is resonant with a homogeneous solution, so use $ax{e}^{-x}$ for its forcing; substitution gives $a=1$.

Thus $y=C_1{e}^{2x}+C_2{e}^{-x}+2e^{-2x}+x{e}^{-x}$. The two initial conditions reduce to $C_1+C_2=3$ and $2C_1-C_2=0$, so $C_1=1$, $C_2=2$. The solution of the initial value problem isThe extra factor $x$ in the resonant particular solution is essential.

The Forward Euler method uses the current slope:Here $x_n=x_0+nh$ and the initial approximation normally equals the given initial value. Each step takes a tangent-line increment of width $h$.

Define the one-step local truncation error as the defect obtained by inserting the exact solution into an update:For a smooth solution, Taylor's theorem and $y^{\prime }=f(x,y)$ giveThus $m=2$ for the one-step defect convention. The leading derivative can also be written $y^{\prime \prime }=f_x+f_{y}f$ by the chain rule.

Some conventions instead call ${\tau }_{n+1}={\delta }_{n+1}/h$ the truncation error. That normalized local truncation error is $O(h)$, so $m=1$ under the normalized convention. Both describe the same Euler one-step defect; neither should be confused with an unqualified global error claim.

Factor the right side as $(x-1{)}^2(x-3)$. On $-1<x<1$, division by $x^2-1$ gives a first-order linear differential equation with integrating factorMultiplication makes the equation $(\mu y{)}^{\prime }=3-x$, so $\mu y=3x-x^2/2+C$. At zero, $\mu (0)y(0)=1$, hence $C=1$. Simplifying givesIts derivative isThe derivation solves the initial-value problem on $(-1,1)$. The simplified expression also extends smoothly through $x=1$ and satisfies the original undivided equation there, giving a solution on all $x>-1$; its pole prevents continuation through $x=-1$.