Consider (x^2-1)y'' + l[x-(α+β)√(x^2-1)r]y' + αβ y = 0, x>1…

Consider $(x^2-1)y'' + \bigl[x-(\alpha+\beta)\sqrt{x^2-1}\bigr]y' + \alpha\beta y = 0, \qquad x>1$. 1° Using the substitution $x=\cosh t$ ($t>0$), reduce the equation to one with constant coefficients and write down its general solution. 2° For $\alpha=5$, $\beta=3$, find the solution that behaves like $x^3$ as $x\to\infty$. 3° Determine the conditions on $\alpha,\beta$ under which every solution stays bounded as $x\to\infty$.

Step 1: The change of variable $x=\cosh t$

For $x>1$ set $x=\cosh t$ with $t>0$, so that

\[ \frac{dx}{dt}=\sinh t, \qquad x^2-1=\cosh^2t-1=\sinh^2t, \qquad \sqrt{x^2-1}=\sinh t. \]

Writing $\dot y=dy/dt$, the chain rule gives $y'=dy/dx=\dot y/\sinh t$, and differentiating again, \[y''=\frac{d}{dx}\left(\frac{\dot y}{\sinh t}\right)=\frac{1}{\sinh t}\frac{d}{dt}\left(\frac{\dot y}{\sinh t}\right)=\frac{\ddot y}{\sinh^2t}-\frac{\cosh t}{\sinh^3t}\,\dot y.\]

Step 2: Substitution into the equation

Insert $x=\cosh t$, $x^2-1=\sinh^2t$, $\sqrt{x^2-1}=\sinh t$, and the two derivative formulas above into the given equation: \[\sinh^2t\left(\frac{\ddot y}{\sinh^2t}-\frac{\cosh t}{\sinh^3t}\dot y\right) + \bigl[\cosh t-(\alpha+\beta)\sinh t\bigr]\frac{\dot y}{\sinh t} + \alpha\beta y = 0.\]

The first bracket contributes $\ddot y-\dfrac{\cosh t}{\sinh t}\dot y$, while the second contributes $\dfrac{\cosh t}{\sinh t}\dot y-(\alpha+\beta)\dot y$; the two terms $\mp\dfrac{\cosh t}{\sinh t}\dot y$ cancel exactly, leaving the constant-coefficient equation

\[ \ddot y - (\alpha+\beta)\dot y + \alpha\beta y = 0, \]

with characteristic equation $r^2-(\alpha+\beta)r+\alpha\beta=(r-\alpha)(r-\beta)=0$.

Step 3: General solution, back in the variable $x$

Since $t=\operatorname{arcosh}x=\ln\bigl(x+\sqrt{x^2-1}\bigr)$, we have $e^t=x+\sqrt{x^2-1}$.

\[y = C_1\bigl(x+\sqrt{x^2-1}\bigr)^{\alpha} + C_2\bigl(x+\sqrt{x^2-1}\bigr)^{\beta} \qquad (\alpha\neq\beta)\]

Step 4: Part 2°: the solution behaving like $x^3$

With $\alpha=5$, $\beta=3$, the general solution is

\[ y = C_1\bigl(x+\sqrt{x^2-1}\bigr)^5 + C_2\bigl(x+\sqrt{x^2-1}\bigr)^3. \]

As $x\to\infty$, $\sqrt{x^2-1}\sim x$, so $x+\sqrt{x^2-1}\sim2x$, and the two terms behave like $32C_1x^5$ and $8C_2x^3$. For $y\sim x^3$ the faster-growing term must be absent, forcing $C_1=0$, and then $8C_2=1$ gives $C_2=\tfrac18$.

\[y = \frac18\bigl(x+\sqrt{x^2-1}\bigr)^3\]

Step 5: Part 3°: boundedness as $x\to\infty$

Since $x+\sqrt{x^2-1}\sim2x\to\infty$, the power $\bigl(x+\sqrt{x^2-1}\bigr)^{\gamma}$ tends to $\infty$ if $\gamma>0$, to $1$ if $\gamma=0$, and to $0$ if $\gamma\lt 0$. For the general solution — with $C_1,C_2$ arbitrary — to remain bounded as $x\to\infty$, neither exponential factor may blow up, so both exponents must be non-positive.

\[\alpha\le0 \quad\text{and}\quad \beta\le0\]