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The characteristic equation has roots $-2,-6$, and a trial multiple of $e^{-4t}$ gives the particular solution $-3e^{-4t}$. ThusThe first two conditions give $A+B=3$ and $-2A-6B+12=0$, hence $A=B=3/2$. At $t_{\ast }=\log \sqrt{2}$, the exponentials $e^{-2t_{\ast }},e^{-4t_{\ast }},e^{-6t_{\ast }}$ equal $1/2,1/4,1/8$. ThereforeNo solution satisfies all three conditions. The first two initial data already determine a unique solution of the second-order linear differential equation; the extra condition is inconsistent with it.

The first equation gives $v=u^{\prime }+3u$. Differentiate it and eliminate $v^{\prime }$ using the second equation:Thus the required second-order linear differential equation isConversely, a solution of this equation together with $v=u^{\prime }+3u$ solves both original equations, so the elimination loses no solutions.

Differentiate $v^{\prime }=u$ once, then substitute the other equation. The required second-order linear differential equation is

Differentiate the second ordinary differential equation and use the first to eliminate $x^{\prime }$:Thus the reduced second-order linear differential equation isNo solutions are lost: given a solution $y$, define $x=y^{\prime }-\cos t$; then both original ordinary differential equations hold.

For an exponential trial $y=e^{rx}$, the characteristic equation isIts distinct complex roots are $r=-a\pm ib$. Thus the general real solution of the homogeneous second-order linear differential equation isThe two displayed solutions have nonzero Wronskian $b{e}^{-2ax}$ and hence linear independence. The same formula permits complex solutions when $C_1,C_2$ are complex.

First simplify the coefficient of $y$ usingAfter dividing by $x(x-m)$, the second-order linear differential equation isAt zero, $xP\to 1$ and $x^{2}Q\to 0$, and both products are analytic there. At $m$, $(x-m)P\to 1$ and $(x-m{)}^{2}Q\to m^2{\omega }^2$, again with analytic extensions. Thus $\boxed{x=0\text{ and }x=m\text{ are regular singular points}}$, and there are no other finite singular points.

Infinity must also be considered because the question asks for all singular points. Put $t=1/x$ and $Y(t)=y(1/t)$. The coefficient of $Y$ in its normal-form equation is $Q(1/t)/t^4\sim {\omega }^2/t^4$. Multiplication by $t^2$ does not make it analytic at zero. Hence $\boxed{x=\infty \text{ is an irregular singular point}}$.

A second-order ordinary differential equation involves a second derivative as its highest derivative. In explicit form $y^{\prime \prime }=f(x,y,y^{\prime })$, introducing $v=y^{\prime }$ converts it into the first-order system $y^{\prime }=v$, $v^{\prime }=f(x,y,v)$. Under local regularity and Lipschitz continuity in the dependent variables, an initial value problem is specified by both $y(x_0)$ and $y^{\prime }(x_0)$. The linear case is a second-order linear differential equation.

Two differentiable functions can be linearly independent on a connected interval while their Wronskian vanishes everywhere. For example, let $f(x)=(\max \{-x,0\}{)}^2$ and $g(x)=(\max \{x,0\}{)}^2$. Both have a continuous first derivative, and their values and first derivatives vanish at zero. On either side of zero one of the functions is identically zero, so $f{g}^{\prime }-f^{\prime }g=0$ everywhere. Evaluating a constant linear combination $Af+Bg=0$ at $-1$ and $1$ forces $A=B=0$.

The obstruction is that the two halves supply independent information without simultaneous nonzero value-and-derivative data at any point. For solutions of a common regular second-order linear differential equation, the uniqueness theorem for ordinary differential equations removes this possibility; Abel's identity gives the zero-or-nowhere-zero alternative for their Wronskian.