Solving a differential equation via reduction of order I want to solve the following differential equation
$$ u^{''}(t)+\frac{2(2-t)}{t(1-t)} u'(t)+\frac{2(1+t)}{t^2(1-t)} u(t)=0, \ t \in (0,1).$$
In order to find the general solution I have to find two linearly independent solutions. I tried to determine the first solution and then use reduction of order to find the second solution. For the first solution I used the ansatz $u(t)=t^{\alpha}, \ \alpha \in R$. Inserting this into the equation leads to
$$ 0=\alpha (\alpha-1) t^{\alpha-2}+\frac{2(2-t)}{t(1-t)} \alpha \ t^{\alpha-1}+\frac{2(1+t)}{t^2(1-t)} t^{\alpha} \\
=\alpha (\alpha-1) t^{\alpha-2} + \frac{2(2-t)}{(1-t)} \alpha \ t^{\alpha-2}
+\frac{2(1+t)}{(1-t)} t^{\alpha-2}$$
Dividing by $t^{\alpha-2}$ leads to
$$ 0=\alpha (\alpha-1) + \frac{2(2-t)}{(1-t)} \alpha
+\frac{2(1+t)}{(1-t)} \\
=\alpha^2+\left(\frac{2(2-t)}{(1-t)}-1 \right) \alpha+\frac{2(1+t)}{(1-t)} \\
=\alpha^2+\frac{3-t}{1-t} \alpha+\frac{2(1+t)}{(1-t)}.$$
Now, in order to find the first solution I need to determine $\alpha \in R$. However, my attempts to solve the equation above have lead to solutions for $\alpha$ that depend on $t \in (0,1)$ which doesn't seem right. What am I missing here?
 A: Taking partial fraction of each term we have
$u''(t)+\left(\dfrac{4}{t}-\dfrac{2}{t-1}\right)u'(t)+\left(\dfrac{2}{t^2}+\dfrac{4}{t}-\dfrac{4}{t-1}\right)u(t)=0$
Let $u(t)=t^kv(t)$ ,
Then $u'(t)=t^kv'(t)+kt^{k-1}v(t)$
$u''(t)=t^kv''(t)+kt^{k-1}v'(t)+kt^{k-1}v'(t)+k(k-1)t^{k-2}v(t)=t^kv''(t)+2kt^{k-1}v'(t)+k(k-1)t^{k-2}v(t)$
$\therefore t^kv''(t)+2kt^{k-1}v'(t)+k(k-1)t^{k-2}v(t)+\left(\dfrac{4}{t}-\dfrac{2}{t-1}\right)(t^kv'(t)+kt^{k-1}v(t))+\left(\dfrac{2}{t^2}+\dfrac{4}{t}-\dfrac{4}{t-1}\right)t^kv(t)=0$
$v''(t)+\dfrac{2k}{t}v'(t)+\dfrac{k(k-1)}{t^2}v(t)+\left(\dfrac{4}{t}-\dfrac{2}{t-1}\right)v'(t)+\left(\dfrac{4k}{t^2}-\dfrac{2k}{t(t-1)}\right)v(t)+\left(\dfrac{2}{t^2}+\dfrac{4}{t}-\dfrac{4}{t-1}\right)v(t)=0$
$v''(t)+\left(\dfrac{2k+4}{t}-\dfrac{2}{t-1}\right)v'(t)+\left(\dfrac{k^2+3k+2}{t^2}+\dfrac{2k+4}{t}-\dfrac{2k+4}{t-1}\right)v(t)=0$
Choosing $k=-2$ , the ODE becomes
$v''(t)-\dfrac{2}{t-1}v'(t)=0$
$\dfrac{v''(t)}{v'(t)}=\dfrac{2}{t-1}$
$\ln v'(t)=2\ln(t-1)+c$
$v'(t)=C(t-1)^2$
$v(t)=C_1(t-1)^3+C_2$
$\therefore u(t)=\dfrac{C_1(t-1)^3}{t^2}+\dfrac{C_2}{t^2}$
