Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 4

At an ordinary point, an order-$n$ linear ordinary differential equation has $n$ arbitrary constants in its general solution. Determining them uniquely requires $n$ independent, consistent scalar conditions. Prescribing $y,y^{\prime },\dots ,y^{(n-1)}$ at one ordinary point gives uniqueness under the standard continuous-coefficient assumptions. Conditions at different endpoints need an additional nonsingularity check: for example, $y^{\prime \prime }+y=0$, $y(0)=y(\pi )=0$ leaves every multiple of $\sin x$ possible. Thus counting boundary conditions alone does not guarantee uniqueness for a boundary value problem.

The characteristic equation is $m^2+3m+2=(m+1)(m+2)=0$, giving the complementary function $C_1{e}^{-x}+C_2{e}^{-2x}$. For the forcing term $x$, a linear particular solution $ax+b$ gives $2ax+3a+2b=x$, hence $a=1/2$, $b=-3/4$.

The forcing $-e^{-x}$ gives resonance in a differential equation, since $e^{-x}$ already appears in the complementary function. Try $Kx{e}^{-x}$. Direct differentiation givesso $K=-1$. Combining the two particular solutions givesThe factor $x$ in the resonant exponential particular solution is required because $-1$ is already a root of the characteristic equation.

Now the characteristic equation is $(m+1{)}^2=0$, with a repeated root, so the complementary function is $(C_1+C_{2}x)e^{-x}$. For the polynomial forcing, substitute $a{x}^2+bx+c$ to obtainThus $a=1$, $b=-4$, $c=6$. For the forcing $2e^{-x}$, both $e^{-x}$ and $x{e}^{-x}$ already occur in the complementary function; the resonant exponential particular solution therefore starts with $K{x}^2{e}^{-x}$. The identitygives $K=1$. The resulting general solution is