Find the homogeneous linear ODE of smallest order having the…

Find the homogeneous linear ODE of smallest order having the given functions as particular solutions $\begin{aligned} 1^\circ&\quad y_1=e^x,\quad y_2=\sinh x,\quad y_3=\cosh x\\[2pt] 2^\circ&\quad y_1=x,\quad y_2=x^3,\quad y_3=|x^3|\\[2pt] 3^\circ&\quad y_1=e^x,\quad y_2=e^x\sin x,\quad y_3=e^x\cos x \end{aligned}$. Method: the Wronskian condition $W(y_1,\dots,y_n,y)=0$, or — when the given set turns out to be dependent, as in $1^\circ$, or genuinely singular, as in $2^\circ$ — a preliminary reduction of the solution set before the ODE is formed.

Step 1: Part 1°  $y_1=e^x,\ y_2=\sinh x,\ y_3=\cosh x$

Since $\sinh x=\tfrac12(e^x-e^{-x})$ and $\cosh x=\tfrac12(e^x+e^{-x})$, both are linear combinations of $e^x$ and $e^{-x}$; conversely, $\cosh x-\sinh x=e^{-x}$, so $e^{-x}$ lies in $\operatorname{span}\{y_1,y_2,y_3\}$ as well. Hence

\[ \operatorname{span}\{e^x,\sinh x,\cosh x\}=\operatorname{span}\{e^x,e^{-x}\}, \] a space of dimension 2, not 3 — the three given functions are linearly dependent, since $\cosh x+\sinh x=e^x$ gives the relation $e^x-\cosh x-\sinh x\equiv0$. The minimal solution space is spanned by $e^x,e^{-x}$, i.e. by characteristic roots $r=1,-1$:

\[ (r-1)(r+1)=r^2-1=0. \] $y'' - y = 0$.

Check: $(e^x)''-e^x=0$; $(\sinh x)''-\sinh x=\sinh x-\sinh x=0$; $(\cosh x)''-\cosh x=\cosh x-\cosh x=0$. $\checkmark$

Step 2: Part 2°  $y_1=x,\ y_2=x^3,\ y_3=|x^3|$

As in Problem 66, $\{x,\,x^3,\,|x^3|\}$ is a linearly independent set of functions on all of $\mathbb{R}$: if $a\,x+b\,x^3+c\,|x^3|\equiv0$, then taking $x>0$ (where $|x^3|=x^3$) forces $a=0,\ b+c=0$, while taking $x<0$ (where $|x^3|=-x^3$) forces $a=0,\ b-c=0$; together these give $a=b=c=0$. So no constant-coefficient linear relation among the three holds on the whole line, even though their Wronskian vanishes identically (a genuine subtlety, not an error — the Wronskian test for dependence presumes enough differentiability, which fails here).

The obstruction is that $y_3=|x^3|$ is not three times differentiable at $x=0$: writing $f(x)=|x|^3$, one finds $f''(x)=6x\,\mathrm{sgn}(x)\to0$ from both sides (so $f''(0)=0$ exists), but $f'''$ jumps from $-6$ to $+6$ across $x=0$ and hence does not exist there. A genuine third-order ODE with continuous coefficients cannot have $y_3$ as a classical solution at $x=0$; and since $\{x,x^3,|x^3|\}$ spans a 3-dimensional space, no second-order equation (2-dimensional solution space) can hold all three either. So no single finite-order linear ODE with continuous coefficients on all of $\mathbb{R}$ admits all three as solutions.

What can be done is to form the minimal-order ODE valid on each half-line separately, where $|x^3|$ coincides (up to sign) with $x^3$ and the independent set effectively reduces to $\{x,x^3\}$:

\[ \begin{vmatrix} x & x^3 & y \\ 1 & 3x^2 & y' \\ 0 & 6x & y'' \end{vmatrix} = 0. \]

Expanding along the first column: $x\bigl(3x^2y''-6xy'\bigr)-1\bigl(x^3y''-6xy\bigr)=0$.

\[ 3x^3y''-6x^2y'-x^3y''+6xy=0\ \Longrightarrow\ 2x^3y''-6x^2y'+6xy=0. \]

Dividing by $2x$ (valid for $x\neq0$):

$x^2\,y'' - 3x\,y' + 3y = 0 \qquad (x\neq0)$.

On $x>0$, $y_3=|x^3|=x^3=y_2$ redundantly solves this equation; on $x<0$, $y_3=|x^3|=-x^3=-y_2$ does too. So this second-order equation governs $y_1,y_2,y_3$ on each half-line, but not across $x=0$ as a single classical ODE.

Check: $y_1=x$: $x^2\cdot0-3x\cdot1+3x=0$. $y_2=x^3$: $x^2\cdot6x-3x\cdot3x^2+3x^3=6x^3-9x^3+3x^3=0$. $\checkmark$

Step 3: Part 3°  $y_1=e^x,\ y_2=e^x\sin x,\ y_3=e^x\cos x$

These three functions are linearly independent and correspond to the characteristic roots $r_1=1$ (from $y_1$) and $r_{2,3}=1\pm i$ (from $y_2,y_3$). The characteristic equation is

\[ (r-1)\bigl[(r-1)^2+1\bigr]=(r-1)(r^2-2r+2)=0. \]

Expanding: \[(r-1)(r^2-2r+2)=r^3-2r^2+2r-r^2+2r-2=r^3-3r^2+4r-2=0.\] $y''' - 3y'' + 4y' - 2y = 0$.

Check: for $y_1=e^x$: $e^x-3e^x+4e^x-2e^x=0$. For $y_2=e^x\sin x$, with

\[ y_2'=e^x(\sin x+\cos x),\qquad y_2''=2e^x\cos x,\qquad y_2'''=2e^x(\cos x-\sin x), \]

substitution gives

\[ y_2'''-3y_2''+4y_2'-2y_2 = e^x\Bigl[2(\cos x-\sin x)-6\cos x+4(\sin x+\cos x)-2\sin x\Bigr]=0, \]

since the $\cos x$ coefficients sum to $2-6+4=0$ and the $\sin x$ coefficients sum to $-2+4-2=0$. By the identical computation with sine and cosine interchanged, $y_3=e^x\cos x$ also satisfies the equation. $\checkmark$