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 $\begin{aligned} 1^\circ&\quad y_1=\sin x,\quad y_2=\cos x\\[2pt] 2^\circ&\quad y_1=e^x,\quad y_2=xe^x\\[2pt] 3^\circ&\quad y_1=x,\quad y_2=x^2\\[2pt] 4^\circ&\quad y_1=3x,\quad y_2=x-2,\quad y_3=e^x+1 \end{aligned}$. Method: the given functions must satisfy the Wronskian condition $W(y_1,\dots,y_n,y)=0$, which for elementary building blocks (exponentials, powers, trig) can be matched directly to a set of characteristic roots.

Step 1: Part 1°  $y_1=\sin x,\ y_2=\cos x$

Both functions are annihilated by the same second-order operator: $(\sin x)''=-\sin x$ and $(\cos x)''=-\cos x$. They are linearly independent (neither is a constant multiple of the other, and their Wronskian $\sin x\cdot(-\sin x)-\cos x\cdot\cos x=-1\neq0$), so the order cannot be lowered below 2.

$y'' + y = 0$.

Check: $(\sin x)''+\sin x=-\sin x+\sin x=0$; $(\cos x)''+\cos x=-\cos x+\cos x=0$.\ \checkmark

Step 2: Part 2°  $y_1=e^x,\ y_2=xe^x$

These two functions are the standard fundamental pair for a double characteristic root $r=1$: the characteristic equation is

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

Check: for $y_1=e^x$: $e^x-2e^x+e^x=0$. For $y_2=xe^x$, with $y_2'=e^x+xe^x$ and $y_2''=2e^x+xe^x$:

\[ (2e^x+xe^x)-2(e^x+xe^x)+xe^x = (2-2)e^x+(1-2+1)xe^x = 0.\ \checkmark \]

Step 3: Part 3°  $y_1=x,\ y_2=x^2$

With two prescribed solutions, the ODE for $y$ is the requirement that $y$ be linearly dependent on $y_1,y_2$ — i.e. that the Wronskian of $x,\,x^2,\,y$ vanish: \[W = \begin{vmatrix} x & x^2 & y \\ 1 & 2x & y' \\ 0 & 2 & y'' \end{vmatrix} = 0.\]

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

$x^2\,y'' - 2x\,y' + 2y = 0$.

Check: $y_1=x$: $x^2\cdot0-2x\cdot1+2x=0$. $y_2=x^2$: $x^2\cdot2-2x\cdot2x+2x^2=2x^2-4x^2+2x^2=0$. $\checkmark$

Step 4: Part 4°  $y_1=3x,\ y_2=x-2,\ y_3=e^x+1$

These three functions are linearly independent (only the trivial combination $a\,y_1+b\,y_2+c\,y_3\equiv0$ is possible: matching the coefficient of $e^x$ forces $c=0$, then matching the coefficient of $x$ and the constant term forces $a=b=0$), so a third-order equation is required. Set $W(y_1,y_2,y_3,y)=0$:

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

Expanding along the first column (the two zero entries kill the third and fourth cofactors): \[3x\begin{vmatrix}1&e^x&y'\\0&e^x&y''\\0&e^x&y'''\end{vmatrix} -3\begin{vmatrix}x-2&e^x+1&y\\0&e^x&y''\\0&e^x&y'''\end{vmatrix}=0.\]

Each $3\times3$ determinant reduces (expand along its own first column) to a multiple of $e^x(y'''-y'')$: \[\begin{vmatrix}1&e^x&y'\\0&e^x&y''\\0&e^x&y'''\end{vmatrix}=e^x(y'''-y''), \qquad \begin{vmatrix}x-2&e^x+1&y\\0&e^x&y''\\0&e^x&y'''\end{vmatrix}=(x-2)\,e^x(y'''-y'').\]

Substituting back: \[3x\,e^x(y'''-y'') - 3(x-2)\,e^x(y'''-y'') = 3e^x(y'''-y'')\bigl[x-(x-2)\bigr] = 6e^x(y'''-y'') = 0.\]

Since $e^x\neq0$ for all $x$, this forces $y'''-y''=0$:

$y''' - y'' = 0$.

Check: the general solution of $y'''-y''=0$ (characteristic roots $r=0,0,1$) is $y=C_1+C_2x+C_3e^x$, and indeed each given function is such a combination:

\[ 3x = 0+3x+0\cdot e^x,\qquad x-2=-2+x+0\cdot e^x,\qquad e^x+1=1+0\cdot x+1\cdot e^x.\ \checkmark \]

Directly: $y_1=3x$ and $y_2=x-2$ are linear in $x$, so $y_1''=y_2''=y_1'''=y_2'''=0$; and for $y_3=e^x+1$, $y_3''=y_3'''=e^x$, so $y_3'''-y_3''=0$ as well.