Given that y_1=l(x+√(x^2+1)r)^n is a particular solution…

Given that $y_1=\bigl(x+\sqrt{x^2+1}\bigr)^n$ is a particular solution, find the general solution of $(1+x^2)y'' + xy' - n^2y = 0$.

Step 1: A key identity

Let $\varphi = x+\sqrt{x^2+1}$. Differentiating, \[\varphi' = 1 + \frac{x}{\sqrt{x^2+1}} = \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}} = \frac{\varphi}{\sqrt{x^2+1}}.\]

This self-referential identity — $\varphi'$ is proportional to $\varphi$ itself — is what makes every power $\varphi^n$ easy to differentiate.

Step 2: Verifying $y_1=\varphi^n$

By the chain rule and the identity above, $y_1' = n\varphi^{n-1}\varphi' = \frac{n\varphi^n}{\sqrt{x^2+1}}$.

Differentiating again (product rule on $n\varphi^n(x^2+1)^{-1/2}$, reusing $\varphi'=\varphi/\sqrt{x^2+1}$): $y_1'' = \frac{n^2\varphi^n}{x^2+1} - \frac{nx\varphi^n}{(x^2+1)^{3/2}}$.

Substituting into the left side of the ODE, the $(1+x^2)$ in front of $y_1''$ cancels one power of $(x^2+1)$ in each term: \[(1+x^2)y_1'' = n^2\varphi^n - \frac{nx\varphi^n}{\sqrt{x^2+1}}, \qquad xy_1' = \frac{nx\varphi^n}{\sqrt{x^2+1}}.\]

Adding these two and subtracting $n^2y_1=n^2\varphi^n$, the $x\varphi^n/\sqrt{x^2+1}$ terms cancel and the $n^2\varphi^n$ terms cancel: \[(1+x^2)y_1''+xy_1'-n^2y_1 = n^2\varphi^n - \frac{nx\varphi^n}{\sqrt{x^2+1}} + \frac{nx\varphi^n}{\sqrt{x^2+1}} - n^2\varphi^n = 0.\ \checkmark\]

Step 3: Abel's formula and reduction of order

In standard form $P(x)=x/(1+x^2)$, so Abel's formula gives $W(x)=C\exp\!\left(-\int\frac{x\,dx}{1+x^2}\right)=\dfrac{C}{\sqrt{1+x^2}}$. With $y_2=y_1v$, $W=y_1^2v'$, so (taking $C=1$, $n\neq0$)

\[ \varphi^{2n}v' = \frac{1}{\sqrt{1+x^2}} \;\Longrightarrow\; v' = \frac{1}{\varphi^{2n}\sqrt{1+x^2}}. \]

Step 4: Changing variables to $u=\varphi$

From Step 1, $du = \varphi'\,dx = \dfrac{\varphi}{\sqrt{x^2+1}}dx$, i.e. $\dfrac{dx}{\sqrt{x^2+1}} = \dfrac{du}{u}$. So

\[ v'\,dx = \frac{1}{u^{2n}}\cdot\frac{du}{u} = u^{-2n-1}\,du. \]

Integrating ($n\neq0$): $v = -\frac{u^{-2n}}{2n} + C_2 = -\frac{\varphi^{-2n}}{2n} + C_2$.

Step 5: The second solution

Dropping the additive constant $C_2$ (it only contributes a multiple of $y_1$) and the overall factor $-1/(2n)$ (absorbed into an arbitrary constant later), the new part of $y_2=y_1v$ is

\[ y_2 = \varphi^n\cdot\varphi^{-2n} = \varphi^{-n} = \bigl(x+\sqrt{x^2+1}\bigr)^{-n}. \]

Since $\bigl(x+\sqrt{x^2+1}\bigr)\bigl(\sqrt{x^2+1}-x\bigr) = (x^2+1)-x^2 = 1$, the conjugate radical is the reciprocal of $\varphi$, so equivalently

\[ y_2 = \bigl(\sqrt{x^2+1}-x\bigr)^n. \]

Step 6: Verification of $y_2$ and the Wronskian

Let $\psi=\sqrt{x^2+1}-x=\varphi^{-1}$. Then $\psi' = \dfrac{x}{\sqrt{x^2+1}}-1 = -\dfrac{\psi}{\sqrt{x^2+1}}$, the same identity as for $\varphi$ but with a sign flip. Repeating Step 2's computation with $\varphi\to\psi$, $\varphi'/\varphi\to-\psi'/... $, every occurrence of the cross term $x\varphi^n/\sqrt{x^2+1}$ is replaced by $-x\psi^n/\sqrt{x^2+1}$ with an extra sign from $\psi'$, and the two instances of that cross term (one from $(1+x^2)y_2''$, one from $xy_2'$) again arrive with opposite signs and cancel, exactly as before. So $y_2=\psi^n$ satisfies the ODE identically as well.

For the Wronskian, use $y_1'=n\varphi^n/\sqrt{x^2+1}$ and $y_2'=-n\varphi^{-n}/\sqrt{x^2+1}$ (from $y_2=\varphi^{-n}$): \[W(y_1,y_2) = y_1y_2'-y_1'y_2 = \varphi^n\!\left(\frac{-n\varphi^{-n}}{\sqrt{x^2+1}}\right) - \frac{n\varphi^n}{\sqrt{x^2+1}}\varphi^{-n} = -\frac{2n}{\sqrt{x^2+1}},\]

which is nonzero for $n\neq0$, confirming $y_1,y_2$ are independent, and matches Abel's formula from Step 3 up to the constant absorbed in Step 5.

valid for $n\neq0$.