Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability / 4

Use partial fraction decomposition:Integrating each term gives the antiderivativeThis holds on any interval avoiding $-2,0,2$; the constant may differ between these intervals. Differentiation of the logarithm recovers $1/[4(x^2-4)]$, providing a direct check.

The characteristic equation of a constant-coefficient differential equation here is $r^2-3r+2=(r-1)(r-2)=0$. Its homogeneous solution is $C_1{e}^x+C_2{e}^{2x}$. Because the forcing $e^x$ is already a homogeneous mode, the method of undetermined coefficients requires $y_p=Ax{e}^x$ rather than $A{e}^x$.

Here $y_p^{\prime }=A(x+1)e^x$ and $y_p^{\prime \prime }=A(x+2)e^x$, so $y_p^{\prime \prime }-3y_p^{\prime }+2y_p=-A{e}^x$. Thus $A=-1$, and the general solution is

The gradient vanishes when $4x+3y-1=0$ and $3x+2y-1=0$. Subtracting these equations gives $x+y=0$, so the unique critical point is $\boxed{(x,y)=(1,-1)}$, with $f(1,-1)=0$.

The Hessian matrix isIt is indefinite, so the critical point is a saddle point. Explicitly, around that point the change is exactly $2u^2+3uv+v^2$: it is positive for $v=0,u\ne 0$ and negative for $v=-3u/2,u\ne 0$. Thus arbitrarily close points have both larger and smaller values.

On the unit circle the constraint gradient $(2x,2y)$ never vanishes. The Lagrange multiplier equations areA nonzero solution of the first two equations requiresso $\lambda =9/2$ or $\lambda =-1/2$. Multiplying the first equation by $x$, the second by $y$, and adding shows $2f=2\lambda (x^2+y^2)=2\lambda$. Thus each candidate value of $f$ equals its multiplier.

For $\lambda =9/2$, $y=-x/3$, giving $(x,y)=\pm (3,-1)/\sqrt{10}$. For $\lambda =-1/2$, $y=3x$, giving $(x,y)=\pm (1,3)/\sqrt{10}$. Since a continuous function attains its extrema on the compact set given by the unit circle and these exhaust the Lagrange multiplier candidates,Each $\pm$ changes both coordinates together.