Given the equation (*): y'' = α(x)y' + β(x)y, α(x)≥0, β(x)≥0…

Given the equation $(*)\!: \qquad y'' = \alpha(x)y' + \beta(x)y, \qquad \alpha(x)\geq0,\ \beta(x)\geq0$, find the condition on $\alpha,\beta$ under which $(*)$ admits two linearly independent solutions $y_1,y_2$ with $y_1y_2=1$. Then, for $\alpha=1/x$, find $\beta$ and the general solution.

Step 1: Imposing $y_2=1/y_1$

If $y_1y_2=1$ then $y_2=1/y_1$, and this $y_2$ must itself satisfy $(*)$. Differentiating twice, \[y_2' = -\frac{y_1'}{y_1^2}, \qquad y_2'' = -\frac{y_1''}{y_1^2} + \frac{2(y_1')^2}{y_1^3}.\]

Substituting into $y_2''=\alpha y_2'+\beta y_2$ and using $y_1''=\alpha y_1'+\beta y_1$ (since $y_1$ solves $(*)$) to eliminate $y_1''$: \[-\frac{\alpha y_1'+\beta y_1}{y_1^2} + \frac{2(y_1')^2}{y_1^3} = -\frac{\alpha y_1'}{y_1^2} + \frac{\beta}{y_1}.\]

Step 2: The condition simplifies

The terms $-\alpha y_1'/y_1^2$ appear identically on both sides and cancel, leaving

\[ -\frac{\beta}{y_1} + \frac{2(y_1')^2}{y_1^3} = \frac{\beta}{y_1} \;\Longrightarrow\; \frac{2(y_1')^2}{y_1^3} = \frac{2\beta}{y_1} \;\Longrightarrow\; (y_1')^2 = \beta\,y_1^2. \]

So the reciprocal relation $y_1y_2=1$ is compatible with $(*)$ precisely when $y_1$ satisfies $(y_1')^2=\beta y_1^2$, i.e. $y_1'/y_1=\pm\sqrt\beta$ — a condition purely on $\beta$, independent so far of $\alpha$.

Step 3: Linking $\alpha$ to $\beta$

Write $p:=y_1'/y_1=\pm\sqrt\beta$, so $y_1'=py_1$ and $y_1''=(p'+p^2)y_1$. Substituting into $y_1''=\alpha y_1'+\beta y_1=(\alpha p+\beta)y_1$ gives $p'+p^2=\alpha p+\beta$. Since $p^2=\beta$, this reduces to $p'=\alpha p$, i.e.

\[ \alpha = \frac{p'}{p}. \]

Because $p^2=\beta$, differentiating gives $2pp'=\beta'$, so $p'/p=\beta'/(2p^2)=\beta'/(2\beta)$ — the same expression for either sign of $p$.

\[\alpha = \frac{\beta'}{2\beta}, \qquad \text{equivalently} \qquad \beta' = 2\alpha\beta, \qquad \bigl(\ln\beta\bigr)' = 2\alpha\]

Step 4: Case $\alpha=1/x$: solving for $\beta$

The condition $\beta'=2\alpha\beta$ becomes the separable equation $\beta'=2\beta/x$: \[\frac{d\beta}{\beta} = \frac{2\,dx}{x} \;\Longrightarrow\; \ln\beta = 2\ln x + \mathrm{const} \;\Longrightarrow\; \beta = kx^2.\]

Taking $k=1$ gives $\beta=x^2$, so $(*)$ becomes

\[ y'' - \frac1x y' - x^2y = 0. \]

Step 5: Building $y_1$ and $y_2$

Taking the positive branch $p=\sqrt\beta=x$ (for $x>0$): $y_1'/y_1=x$, so $\ln y_1=x^2/2$, giving

\[ y_1 = e^{x^2/2}, \qquad y_2 = \frac{1}{y_1} = e^{-x^2/2}. \]

Step 6: Verification

For $y_1=e^{x^2/2}$: $y_1'=xe^{x^2/2}$, $y_1''=(1+x^2)e^{x^2/2}$. Substituting into $y''-\tfrac1xy'-x^2y$: \[(1+x^2)e^{x^2/2} - \frac1x\bigl(xe^{x^2/2}\bigr) - x^2e^{x^2/2} = \bigl[(1+x^2)-1-x^2\bigr]e^{x^2/2} = 0.\ \checkmark\]

For $y_2=e^{-x^2/2}$: $y_2'=-xe^{-x^2/2}$, $y_2''=(x^2-1)e^{-x^2/2}$. Substituting: \[(x^2-1)e^{-x^2/2} - \frac1x\bigl(-xe^{-x^2/2}\bigr) - x^2e^{-x^2/2} = \bigl[(x^2-1)+1-x^2\bigr]e^{-x^2/2} = 0.\ \checkmark\]

The reciprocal relation holds exactly, $y_1y_2=e^{x^2/2}e^{-x^2/2}=1\ \checkmark$, and the Wronskian

\[ W = y_1y_2'-y_1'y_2 = e^{x^2/2}\bigl(-xe^{-x^2/2}\bigr) - xe^{x^2/2}\bigl(e^{-x^2/2}\bigr) = -2x \]

is nonzero for $x\neq0$, confirming $y_1,y_2$ are linearly independent.

\[\beta = x^2, \qquad y = C_1e^{x^2/2} + C_2e^{-x^2/2}\]