Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 5

The characteristic equation of the constant-coefficient linear ordinary differential equation iswhose roots of a polynomial are $r=-1\pm i/2$. Taking real and imaginary parts of the corresponding exponential functions gives the real general solution

For a particular solution, try $y_p=a\sin x+b\cos x$. Substitution givesMatching coefficients requires $a-8b=1$ and $8a+b=8$, so $a=1$ and $b=0$. Adding the homogeneous solution from part (i), the general solution is

Put $a=\pi /2$. Substituting $x=-a$ and $x=a$ into part (ii), using $\cos (\pi /4)=\sin (\pi /4)=1/\sqrt{2}$, givesThus $A-B=\sqrt{2}(1+e^{-a})$ and $A+B=\sqrt{2}(1-e^a)$. Solving these two linear equations givesConsequently the solution of the boundary value problem isThe two independent boundary conditions determine $A,B$ uniquely.

Since $x>-1$, divide by $x+1$ to obtain the first-order linear ordinary differential equationAn integrating factor is $\mu (x)=e^{-3\log (x+1)}=(x+1{)}^{-3}$. Multiplication by this factor givesAfter integration, $(x+1{)}^{-3}y=\frac{1}{2}(x+1{)}^2+C$, so

Let $g_1=x^2+y^2+z^2-4$ and $g_2=(x-3{)}^2+y^2+z^2-4$. Subtracting the two constraint equations gives $-6x+9=0$, hence $x=3/2$ and $y^2+z^2=7/4$. The feasible set is therefore a circle.

For Lagrange multipliers $\lambda ,\mu$, the stationarity equation $\nabla f=\lambda \nabla g_1+\mu \nabla g_2$ becomesAt $x=3/2$, the first equation gives $\lambda =\mu$. The second ensures $\lambda +\mu \ne 0$ and $y\ne 0$, so the third gives $z=0$. The constraints then force $y=\pm \sqrt{7}/2$. Both points do satisfy the multiplier equations, taking $\lambda =\mu =1/(4y)$.

There are no missing singular constraint points: on the feasible circle, the two constraint gradients $(3,2y,2z)$ and $(-3,2y,2z)$ are linearly independent, since $y^2+z^2=7/4>0$. Thus the complete list of constrained critical points and critical values isThe bound $|y|\le \sqrt{7}/2$ on the feasible circle shows that the positive value is the global maximum and the negative value is the global minimum.