Solve, in terms of the given functions P(x) and Q(x) : y'''…

Solve, in terms of the given functions $P(x)$ and $Q(x)$: $y''' = \bigl(x^2y''-2xy'+2y\bigr)P(x)+Q(x)$, using the substitution $u=x^2y''-2xy'+2y$.

Step 1: Set up the substitution

Introduce the new unknown $u=u(x)$ by

\[ u = x^2y''-2xy'+2y. \]

The equation to be solved then reads simply $y'''=uP(x)+Q(x)$; the whole strategy is to first find $u(x)$, and only afterwards recover $y(x)$ from the definition of $u$.

Step 2: Differentiate $u$: the key cancellation

Differentiating $u=x^2y''-2xy'+2y$ once with respect to $x$: $u' = \bigl(2xy''+x^2y'''\bigr) + \bigl(-2y'-2xy''\bigr) + 2y'.$

The two $2xy''$ terms cancel, and the $-2y'$ and $+2y'$ terms cancel, leaving only

\[ u' = x^2y''', \qquad\text{i.e.}\qquad y''' = \frac{u'}{x^2}. \] So the third-order relation between $y'''$ and $u$ is, remarkably, purely algebraic once expressed through $u'$ — no trace of $y$ itself survives.

Step 3: A first-order linear equation for $u$

Substituting $y'''=u'/x^2$ into the original equation $y'''=uP(x)+Q(x)$: \[\frac{u'}{x^2} = uP(x)+Q(x) \quad\Longrightarrow\quad u' - x^2P(x)\,u = x^2Q(x).\]

This is a first-order linear ODE in $u(x)$ — the third-order problem has been reduced by two full orders.

Step 4: Solve for $u(x)$

With integrating factor $\mu(x)=\exp\!\left(-\displaystyle\int x^2P(x)\,dx\right)$, the equation $\dfrac{d}{dx}\bigl[\mu u\bigr]=\mu\,x^2Q(x)$ integrates to

\[ u(x) = \frac{1}{\mu(x)}\left[\,C_0+\int \mu(x)\,x^2Q(x)\,dx\,\right], \]

with $C_0$ an arbitrary constant of integration.

Step 5: Recover $y$: a non-homogeneous Euler equation

Once $u(x)$ is known, the definition $u=x^2y''-2xy'+2y$ is itself a second-order ODE for $y$ — an Euler (equidimensional) equation with right-hand side $u(x)$:

\[ x^2y''-2xy'+2y = u(x). \]

Its homogeneous version $x^2y''-2xy'+2y=0$ is solved by $x=e^t$; using the standard identities $xy'=\dot y$, $x^2y''=\ddot y-\dot y$ (dots denoting $d/dt$), \[(\ddot y-\dot y)-2\dot y+2y=0 \implies \ddot y-3\dot y+2y=0 \implies r^2-3r+2=(r-1)(r-2)=0,\]

giving roots $r=1,2$, i.e.\ $y=C_1e^t+C_2e^{2t}=C_1x+C_2x^2$. The fundamental system of the homogeneous Euler equation is therefore

\[ y_1=x, \qquad y_2=x^2. \]

Step 6: Variation of parameters

Writing $x^2y''-2xy'+2y=u(x)$ in standard form $y''-\dfrac{2}{x}y'+\dfrac{2}{x^2}y=\dfrac{u}{x^2}$, the Wronskian of $\{x,x^2\}$ is

\[ W(x,x^2) = x\cdot 2x - x^2\cdot 1 = x^2. \]

Variation of parameters then gives, with $f(x)=u/x^2$, \[C_1'(x) = -\frac{y_2 f}{W} = -\frac{x^2\cdot(u/x^2)}{x^2} = -\frac{u}{x^2}, \qquad C_2'(x) = \frac{y_1 f}{W} = \frac{x\cdot(u/x^2)}{x^2} = \frac{u}{x^3}.\]

Integrating and adding fresh arbitrary constants $C_1^{*},C_2^{*}$ (absorbing the constants of these last two integrations) to the homogeneous terms:

Step 7: General solution

The solution is recorded in the result below.