Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 4

The function is even, so every sine coefficient is zero. Its mean coefficient isFor $n\ge 1$, compute $a_n=(2/\pi ){\int }_0^{\pi }\sin x\cos (nx)\mathrm{d}x$. The product-to-sum identity makes this integral zero for odd $n$ and gives $-4/(\pi (n^2-1))$ for even $n$. Thus the missing Fourier coefficients are $a_n=0$ for odd $n$ and $b_n=0$ for every $n$, with $a_0=4/\pi$. Equivalently,The Fourier series converges to this continuous periodic function; there is no jump value to average.

For $x>0$, substitute $v=y/x$. Then $y^{\prime }=v+x{v}^{\prime }$, and cancellation reduces the differential equation to $v^{\prime }+2v=e^{-2x}$. Multiplication by the integrating factor $e^{2x}$ gives $(e^{2x}v{)}^{\prime }=1$. ThereforeDirect substitution gives $y^{\prime }+2y-y/x=x{e}^{-2x}$ and verifies the result. Dividing by $x$ is legitimate on the stated positive domain.

The integrating factor is $e^{2x}$, so the differential equation becomes $(e^{2x}y{)}^{\prime }=1$. Integration givesThe factor $x$ in the particular solution accounts for the forcing term being proportional to the homogeneous solution.

Here the integrating factor is $e^{2\sin x}$, since its logarithmic derivative is $2\cos x$. Multiplying the differential equation by it cancels the right-hand exponential and yields $(e^{2\sin x}y{)}^{\prime }=1$. HenceThis formula is actually defined for every real $x$.

Seek a periodic particular solution with Fourier coefficients $A_n,B_n$. Comparing constant terms gives $A_0=a_0/2$. Comparing the cosine and sine terms for $n\ge 1$ givesso $A_n=2a_n/(n^2+4)$ and $B_n=n{a}_n/(n^2+4)$. Insert the coefficients from part (a) and add the homogeneous solution:The differentiated series converges absolutely and uniformly, because its largest coefficient terms are $O(m^{-2})$. This justifies the term-by-term first derivative used in substituting into the differential equation. Any two solutions differ by $C{e}^{-2x}$, proving that the displayed family is the general solution. The computation is an instance of Fourier response of a first-order linear differential equation.