Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 20

Seek a separated solution $u=X(x)Y(y)$. On a region where the factors do not vanish, the partial differential equation separates asThe resulting ordinary differential equations integrate to $X=A{e}^{\lambda x}$ and $Y=B|3-y{|}^{\lambda /2}$ on either side of $y=3$. Choose $\lambda =2$ and the signed linear branch $Y=3-y$, which extends smoothly across that line. This yields the nonconstant solutionDirect partial differentiation gives $u_x=2e^{2x}(3-y)$ and $u_y=-e^{2x}$, whose prescribed combination vanishes everywhere. This direct check also covers the line on which the separated factor is zero.

With $u=XY$, divide the separated equation on $x\ne 0$, $y\ne 0$ to obtainThus $X^{\prime }/X=\lambda x{e}^{-x}$ and $Y^{\prime }/Y=\lambda /y$. Taking antiderivatives,Choose $\lambda =1$ and the signed branch $Y=y$. A nonconstant solution, smooth for every real $x,y$, isIndeed $u_x=x{e}^{-x}u$ and $u_y=\exp [-(x+1)e^{-x}]$, so $e^x{u}_x-xy{u}_y=0$. This verification includes the axes excluded temporarily during separation.

For $u=XY$, the separated equations areOn a region containing the prescribed points, take $0<x<\pi$ and $0<y<\pi /2$. Integration givesThe two specified values, verified in the original PDF, giveDividing these relations eliminates $C$:The right-hand side is the square of the positive base on the left, which differs from one. Thus $\lambda =2$, and either original condition gives $C=1$. ThereforeIts partial derivatives satisfy $u_x/\sin x=2u$ and $\tan y{u}_y=2u$, verifying the equation wherever its coefficients are defined. Substitution at the two assigned points gives $1/(4e^{\sqrt{3}})$ and $3/(4e)$ exactly. The formula extends to any region where $\sin x\ne 0$ and $\cos y\ne 0$.