Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 5A (Differential Equations)

Let $D=d/dx$ and write the constant-coefficient linear ordinary differential equation as $p(D)y=0$. Since $D^k{e}^{\lambda x}={\lambda }^k{e}^{\lambda x}$,The exponential never vanishes, so $\boxed{e^{\lambda x}\text{ solves the equation }⟺p(\lambda )=0}$.

For the second claim, the product rule givesfor $k\ge 1$, with the $k=0$ term treated separately. Summing the coefficients yieldsA root of multiplicity at least two has $p(\mu )=p^{\prime }(\mu )=0$, proving $x{e}^{\mu x}$ is also a solution. This directly establishes the repeated-root contribution without assuming a solution formula. If the displayed coefficient list has leading zeros, the exponential calculation still applies to its actual polynomial degree.

Set $v=u^{\prime \prime }$. The resulting second-order linear differential equation is $v^{\prime \prime }+2v=4t^2$. A polynomial particular solution $v_p=a{t}^2+b$ requires $2a=4$ and $2a+2b=0$, hence $v_p=2t^2-2$. The homogeneous characteristic roots of a constant-coefficient differential equation are $\pm i\sqrt{2}$, soIntegrating twice and renaming constants gives the general real solutionThe four independent homogeneous terms account for the four initial data of the fourth-order linear ordinary differential equation.

For the substitution $u(t)=y(t^2)$, repeated use of the chain rule, with $x=t^2$, givesThereforeFor $x>0$, the map $t=\sqrt{x}$ is invertible, and the transformed equation is exactly the equation just solved. Hence the general real solution on $x>0$ isThere is no requirement that $u$ extend as an even function through $t=0$: only the branch $t>0$ is needed, so the sine and square-root terms must be retained.