Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 17 / b

In the homogeneous equation, substitute the exponential function $e^{kx}$. The characteristic equation isIts distinct roots are $0,1,-1,i,-i$. Combining the complex-conjugate pair into real sine and cosine gives the complementary functionFive linearly independent terms supply the required five homogeneous degrees of freedom.

A quadratic polynomial is sufficient: set $y_p=A{x}^2$. Its fifth derivative vanishes, so $y_p^{(5)}-y_p^{\prime }=-2Ax$. Matching the forcing gives $A=-1/2$, hence a particular integral is

Add the particular solution to the complementary function. The real general solution isThe five constants remain arbitrary until initial data are specified.

Impose the five initial conditions at zero on the preceding general solution. They giveThe second and fourth equations give $C_1=C_2$ and $C_4=0$. The third and fifth then yield $C_1=C_2=1/4$ and $C_3=-1/2$, while the first gives $C_0=1$. ThusFor a direct check, the difference of its fifth and first derivatives is $x$. At zero its value is one; its first, second, third and fourth derivatives vanish. In particular the quadratic term cancels the otherwise nonzero second derivative of the hyperbolic-cosine and cosine combination.