Consider y'' + P(x)y' + Q(x)y = 0 . 1° Find the condition on…

Consider $y'' + P(x)y' + Q(x)y = 0$. 1° Find the condition on $P(x)$ and $Q(x)$ under which a change of independent variable $t=f(x)$ reduces this equation to one with constant coefficients, $\ddot y + a\dot y + by = 0$. 2° Apply this to $P(x)=2/x$: find the corresponding $Q(x)$, and solve the resulting constant-coefficient equation.

Step 1: Transforming the equation under $t=f(x)$

Regard $y$ as a function of the new variable $t=f(x)$, and write dots for $d/dt$. By the chain rule, \[\frac{dy}{dx} = f'(x)\,\dot y, \qquad \frac{d^2y}{dx^2} = \frac{d}{dx}\big[f'(x)\dot y\big] = f'(x)^2\ddot y + f''(x)\dot y.\]

Substituting into $y''+Py'+Qy=0$ gives

\[ f'(x)^2\ddot y + f''(x)\dot y + P(x)f'(x)\dot y + Q(x)y = 0, \]

and dividing through by $f'(x)^2$ puts this in the form

\[ \ddot y + \left[\frac{f''(x)}{f'(x)^2} + \frac{P(x)}{f'(x)}\right]\dot y + \frac{Q(x)}{f'(x)^2}\,y = 0. \]

Step 2: Demanding constant coefficients

For this to match a target equation $\ddot y + a\dot y + by=0$ with prescribed constants $a,b$, both bracketed coefficients must be constant. The $y$-coefficient fixes $f'(x)$ outright:

\[ \frac{Q(x)}{f'(x)^2} = b \quad\Longrightarrow\quad f'(x) = \sqrt{\frac{Q(x)}{b}}. \]

The $\dot y$-coefficient then imposes a second condition, relating $P$ to $Q$: $\frac{f''(x)}{f'(x)^2} + \frac{P(x)}{f'(x)} = a$.

Differentiating $f'=\sqrt{Q/b}$ gives $f''=\dfrac{Q'}{2\sqrt{b}\sqrt{Q}}$, so that

\[ \frac{f''}{f'^2} = \frac{Q'/(2\sqrt bQ^{1/2})}{Q/b} = \frac{\sqrt b\,Q'}{2Q^{3/2}}, \qquad \frac{P}{f'} = P\sqrt{\frac{b}{Q}} = \frac{P\sqrt b}{\sqrt Q}. \]

Adding these and factoring out $\sqrt b$ produces the clean closed-form condition on $P$ and $Q$ alone: \[\ \sqrt{b}\,\frac{P(x)Q(x)+\tfrac12 Q'(x)}{Q(x)^{3/2}} = a \quad\Longleftrightarrow\quad \sqrt{\frac{b}{Q(x)}}\left[P(x) + \frac{Q'(x)}{2Q(x)}\right] = a\\]

for the prescribed constants $a,b$. Whenever $P$ and $Q$ satisfy this relation, the substitution $t=f(x)=\displaystyle\int\sqrt{Q(x)/b}\,dx$ carries $y''+Py'+Qy=0$ into the constant-coefficient equation $\ddot y+a\dot y+by=0$.

Step 3: Applying the condition to $P(x)=2/x$

With $P=2/x$, the boxed condition becomes a first-order ODE for $Q(x)$: \[\sqrt{\frac{b}{Q}}\left[\frac{2}{x} + \frac{Q'}{2Q}\right] = a.\]

This is of Bernoulli type in $Q$; it linearizes under the substitution $R(x) = Q(x)^{-1/2}$, so that $Q=R^{-2}$ and $Q'=-2R^{-3}R'$. Then

\[ \sqrt{\frac{b}{Q}} = \sqrt b\,R, \qquad \frac{Q'}{2Q} = \frac{-2R^{-3}R'}{2R^{-2}} = -\frac{R'}{R}, \]

and the condition becomes $\sqrt b\,R\big[\tfrac2x - \tfrac{R'}{R}\big]=a$, i.e.

\[ \sqrt b\left(\frac{2R}{x} - R'\right) = a \quad\Longrightarrow\quad R'(x) - \frac{2R(x)}{x} = -\frac{a}{\sqrt b}. \]

This is now a linear first-order equation. Its integrating factor is $\mu(x)=x^{-2}$, since $\dfrac{d}{dx}\big(Rx^{-2}\big) = x^{-2}R' - 2x^{-3}R = x^{-2}\big(R'-2R/x\big)$. Multiplying through by $x^{-2}$ and integrating,

\[ \frac{d}{dx}\big(Rx^{-2}\big) = -\frac{a}{\sqrt b}\,x^{-2} \quad\Longrightarrow\quad Rx^{-2} = \frac{a}{\sqrt b}\,x^{-1} + C, \]

so that, writing $\alpha := a/\sqrt b$ for brevity, $R(x) = Cx^2 + \alpha x$.

Undoing $R=Q^{-1/2}$ gives the general family of admissible $Q$'s: \[\ Q(x) = \frac{1}{x^2\big(Cx+\alpha\big)^2}\ }, \qquad \alpha = \frac{a}{\sqrt b},\ C \text{ arbitrary\]

(Substituting this $Q$ back into the boxed condition of Part 1° reduces it identically to $0=0$, confirming the family is exactly right, on whatever interval keeps $Q$ positive and $x\ne0$, $Cx+\alpha\ne0$.)

Step 4: The cleanest special case: $\alpha=0$

Take $a=0$ (so $\alpha=0$) and choose the free constant as $C=1/\sqrt b$ for a given $b>0$. Then $R(x)=x^2/\sqrt b$ and

\[ Q(x) = \frac{1}{R(x)^2} = \frac{b}{x^4}. \]

So for $P(x)=2/x$ and $Q(x)=b/x^4$, the theory guarantees a reduction to $\ddot y + by = 0$ (since $a=0$ here). The change of variable is $f'(x)=\sqrt{Q/b}=\sqrt{1/x^4}=1/x^2$, so

\[ t = f(x) = -\frac{1}{x}. \]

A direct check confirms this works: with $t=-1/x$, so $dt/dx=1/x^2$, the chain rule gives $y'=\dot y/x^2$ and $y''=\ddot y/x^4 - 2\dot y/x^3$. Substituting into $y''+\tfrac2x y'+\tfrac{b}{x^4}y=0$,

\[ \frac{\ddot y}{x^4} - \frac{2\dot y}{x^3} + \frac{2}{x}\cdot\frac{\dot y}{x^2} + \frac{b}{x^4}y = \frac{\ddot y}{x^4} + \frac{b}{x^4}y = \frac{1}{x^4}\big(\ddot y + by\big) = 0, \]

and since $x^4\ne0$ this is exactly $\ddot y + by = 0$, the promised constant-coefficient equation.

Step 5: Solving and returning to $x$

The equation $\ddot y+by=0$ has general solution (with $b>0$)

\[ Y(t) = C_1\cos\sqrt b\,t + C_2\sin\sqrt b\,t. \]

Substituting $t=-1/x$, and using that cosine is even while sine is odd — so the minus sign only rescales $C_2$, which is arbitrary anyway — gives the general solution back in terms of $x$:

\[y(x) = C_1\cos\frac{\sqrt b}{x} + C_2\sin\frac{\sqrt b}{x}\]

valid for $P(x)=2/x$, $Q(x)=b/x^4$, on any interval avoiding $x=0$.