Find the general solution of (1+x^2)y'' + xy' - n^2y = 0…

Find the general solution of $(1+x^2)y'' + xy' - n^2y = 0$, given that $y_1 = \left(x+\sqrt{x^2+1}\right)^n$ is a particular solution.

Step 1: Properties of $\varphi=x+\sqrt{x^2+1}$

Write $y_1=\varphi^n$. Since

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

and $\varphi\cdot(\sqrt{x^2+1}-x) = (x^2+1)-x^2=1$, so $\varphi^{-1}=\sqrt{x^2+1}-x$. Then

\[ y_1' = n\varphi^{n-1}\varphi' = \frac{n\varphi^n}{\sqrt{x^2+1}} = \frac{ny_1}{\sqrt{x^2+1}}, \qquad y_1'' = \frac{n^2y_1}{x^2+1}-\frac{nxy_1}{(x^2+1)^{3/2}}. \]

Step 2: Verify $y_1$

\[ (1+x^2)y_1''+xy_1'-n^2y_1 = n^2y_1 - \frac{nxy_1}{\sqrt{x^2+1}} + \frac{nxy_1}{\sqrt{x^2+1}} - n^2y_1 = 0.\ \checkmark \]

Step 3: Substitution $y=\varphi^nv$

\[ y' = \varphi^n\!\left(\frac{nv}{\sqrt{x^2+1}}+v'\right), \qquad y'' = \varphi^n\!\left(\frac{n^2v}{x^2+1}-\frac{nxv}{(x^2+1)^{3/2}}+\frac{2nv'}{\sqrt{x^2+1}}+v''\right). \]

Step 4: Reduce the order

Substitute into $(1+x^2)y''+xy'-n^2y=0$ and divide by $\varphi^n$. The $n^2v$ terms and the $\dfrac{nxv}{\sqrt{x^2+1}}$ terms cancel, leaving

\[ (1+x^2)v'' + \left(2n\sqrt{x^2+1}+x\right)v' = 0. \]

Step 5: Solve for $w=v'$

\[ \frac{dw}{w} = -\frac{2n\sqrt{x^2+1}+x}{1+x^2}\,dx = -\frac{2n}{\sqrt{x^2+1}}\,dx - \frac{x}{1+x^2}\,dx. \]

Using $\displaystyle\int\frac{dx}{\sqrt{x^2+1}}=\ln\varphi$ and $\displaystyle\int\frac{x\,dx}{1+x^2}=\frac12\ln(1+x^2)$: \[\ln|w| = -2n\ln\varphi - \frac12\ln(1+x^2)+\mathrm{const} \implies w = \frac{A}{\varphi^{2n}\sqrt{x^2+1}} = \frac{At^{2n}}{\sqrt{x^2+1}}, \qquad t:=\varphi^{-1}=\sqrt{x^2+1}-x.\]

Step 6: Integrate for $v$

Differentiating $t=\sqrt{x^2+1}-x$ directly (not via $\varphi$): \[\frac{dt}{dx} = \frac{x}{\sqrt{x^2+1}}-1 = \frac{x-\sqrt{x^2+1}}{\sqrt{x^2+1}} = -\frac{t}{\sqrt{x^2+1}}, \qquad\text{so}\qquad \frac{dx}{\sqrt{x^2+1}} = -\frac{dt}{t}.\]

Then \[v = A\int t^{2n}\cdot\frac{dx}{\sqrt{x^2+1}} = A\int t^{2n}\!\left(-\frac{dt}{t}\right) = -A\int t^{2n-1}\,dt = -\frac{A}{2n}t^{2n}+B.\]

Step 7: General solution

Multiply by $y_1=\varphi^n$, and use $\varphi^nt^{2n}=\varphi^n(\varphi^{-1})^{2n}=\varphi^{-n}=t^n$ (since $\varphi t=1$): $y = \varphi^n\!\left[B-\frac{A}{2n}t^{2n}\right] = B\varphi^n - \frac{A}{2n}t^n.$

Renaming $C_1=B$, $C_2=-A/(2n)$:

Step 8: Verify $y_2=\left(\sqrt{x^2+1}-x\right)^n=\varphi^{-n}$

The equation depends on $n$ only through $n^2$, so it is invariant under $n\to-n$. Substituting $-n$ for $n$ in Steps 1–2's identities: \[y_2' = -\frac{ny_2}{\sqrt{x^2+1}}, \qquad y_2'' = \frac{n^2y_2}{x^2+1}+\frac{nxy_2}{(x^2+1)^{3/2}},\]

\[ (1+x^2)y_2''+xy_2'-n^2y_2 = n^2y_2+\frac{nxy_2}{\sqrt{x^2+1}}-\frac{nxy_2}{\sqrt{x^2+1}}-n^2y_2=0.\ \checkmark \]

Step 9: Linear independence

In standard form the coefficient of $y'$ is $P(x)=\dfrac{x}{1+x^2}$, so by Abel's formula

\[ W(y_1,y_2) = W_0\exp\!\left(-\int\frac{x\,dx}{1+x^2}\right) = \frac{W_0}{\sqrt{1+x^2}} \neq 0, \]

confirming that $y_1,y_2$ form a fundamental system.