Past exam of the mathematical methods I course of the University of Oxford / 2021 resit / 10

For a genuinely second-order linear ordinary differential equation, assume $a_2\ne 0$. Substituting the exponential function $f=e^{rt}$ into the homogeneous linear differential equation gives $a_2{r}^2+a_{1}r+a_0=0$. Oscillations arise precisely when its two roots of a polynomial have nonzero imaginary parts, namelyIn that case, writing $\omega =\sqrt{4a_0{a}_2-a_1^2}/(2|a_2|)>0$, the real complementary function isEvery nonzero such solution has an oscillatory factor; the oscillation may decay, grow, or keep constant amplitude according to the sign of $a_1/a_2$. This is the oscillatory-root criterion for a real second-order equation; decay is not a prerequisite for oscillation.

For completeness, if $a_2=0$ and $a_1\ne 0$, only real exponential functions occur. If $a_2=a_1=0$ and $a_0\ne 0$, only $f=0$ is allowed. If all three coefficients vanish, every twice-differentiable function is allowed, including oscillatory ones; the second-order criterion does not apply to this degenerate identity.

Use the method of undetermined coefficients with the polynomial $f_p=a{t}^3+b{t}^2+ct+d$. Substitution into $f_p^{\prime \prime }+3f_p^{\prime }+2f_p$ givesMatching successive coefficients with the forcing gives $a=1$, $b=-3$, $c=4$ and $d=1$. Hence a particular integral isIndeed $f_p^{\prime }=3t^2-6t+4$ and $f_p^{\prime \prime }=6t-6$ give $f_p^{\prime \prime }+3f_p^{\prime }+2f_p=2t^3+3t^2-4t+8$. The complementary function, if wanted, is $C_1{e}^{-t}+C_2{e}^{-2t}$.

The differential operator is $(D-1{)}^2$, with $D=d/dt$. Because $e^t$ already occurs in the complementary function with a repeated root of a polynomial, an unmodified trial $C{e}^t$ gives zero. Put $f_p=e^{t}h(t)$. The product rule yieldsThus $h^{\prime \prime }=2$, and we may choose $h=t^2$, givingThe other terms arising on twice integrating $h^{\prime \prime }=2$ give $(C_1+C_{2}t)e^t$, exactly the complementary function, so they need not be included in the particular integral.

The homogeneous linear differential equation has roots $(-1\pm i\sqrt{3})/2$, so its complementary function decays:For a particular integral, set $f_p=B\cos 2t+C\sin 2t$. Substitution givesThe two coefficient equations are $-3B+2C=A$ and $-2B-3C=0$, so $B=-3A/13$ and $C=2A/13$. Consequently every solution obeysThe limiting behaviour is a periodic steady oscillation with amplitude $|A|/\sqrt{13}$, the long-time response of a forced harmonic oscillator. Unless $A=0$, there is no finite numerical limit: at $t=n\pi$ and $t=n\pi +\pi /2$, the two limiting values are $-3A/13$ and $3A/13$. For $A=0$, $f(t)\to 0$.