Let $\alpha$ and $\beta$ with $\alpha>\beta$ be the roots of the indicial equation of $(x^2-1)^2y''+(x+1)y'-y=0$ at $x= -1$.
Then what is the value of $\alpha-4\beta$ ?
I am trying to solve this by Frobenius series solution method. Assuming trial solution of the form $y=\sum_{m=0}^{\infty} c_m(x+1)^{m+\rho}$. Then I get the indicial equation as $$\rho^2-1=0.$$ So $\alpha=1$ and $\beta=-1$. Thus the value of $\alpha-4\beta$ is $5$.
But the answer is $2$. The question appears in GATE 2017. So where i am wrong. Any help is highly appreciated.
Best Answer
$$(x^2-1)^2y''+(x+1)y'-y=0$$ At $x=-1$ We have, $$\alpha(x)(x+1)^2y''+\beta(x)(x+1)y'+\gamma(x)y=0$$ With$ \begin{cases} \alpha(x)=(x-1)^2 \implies \alpha(-1)=4\\ \beta(x)=1 \implies \beta(-1)=1\\ \gamma(x)=-1 \implies \gamma(-1)=-1 \end{cases} $
Then the indicial equation is
$$r^2+\left(\frac {\beta(-1)}{\alpha(-1)}-1\right)r+\frac {\gamma(-1)}{\alpha(-1)}=0$$ $$r^2-\frac {3}{4}r-\frac {1}{4}=0$$ $$(r-1)(r+\frac {1}{4})=0 \implies r=1, r=-\frac 1 4$$
$$\text{And }\alpha -4\beta=1-4\frac {-1} 4=2$$