Finding and proving linear independence of solutions of second order differential equation Show that the solutions of the differential equation $2x^2y'' + 10xy' + 8y = 0$ are linearly
independent in two ways and find its particular solution satisfying the initial conditions
$y(1) = 3$ and $y'(1) = 5$
I have tried this problem by using power series method but ended up getting only the trivial solution as my answer. Any help would be appreciated.
 A: Your equation is a Cauchy-Euler equation which can be solved by assuming that the solution is of the form $y(x) = x^k$ for some $k \in\mathbb{R}$.
Plugging this in gives you:
$$2x^2(k-1)kx^{k-2} + 10xkx^{k-1} + 8x^k = 0$$
Factoring out $x^k$, we obtain:
$$x^k(2k^2 - 2k + 10k + 8) = x^k(2k^2+8k+8) = 0$$
However, this must hold for any value of $x$, so we know that $2k^2 + 8k + 8 = 0$ must hold, which is equivalent to $k^2 + 4k + 4 = 0$. We can easily solve this with the quadratic formula and obtain $k = -2$ as the only solution.
Because this is a repeated solution to our quadratic, the general solution of the initial ODE is given by
$$y(x) = C_1 y_1 (x) + C_2 y_2(x)=\frac{C_1}{x^2} + \frac{C_2\log(x)}{x^2}$$
Using our first initial condition $y(1) = 3$, we have
$$3 = y(1) = \frac{C_1}{1} + \frac{C_2\log(1)}{1} = C_1$$
Plugging this in and calculating $y'(x)$, we have
$$y'(x) = -\frac{6}{x^3} + C_2\frac{1-2\log(x)}{x^3}$$
and thus, with our second initial condition that $y'(1) = 5$, we obtain
$$5 = y'(1) = -6 + C_2 \Rightarrow C_2 = 11$$
Our particular solution is therefore
$$y(x) = \frac{6}{x^2} + \frac{11\log(x)}{x^2}$$
The linear independence of the two solutions of the general equation is pretty obvious and can be shown directly (by showing that no $C_1, C_2 \in \mathbb{R}$ with at least one of them non-zero exist such that $y(x) =  0$ for all $x$ - this is fairly straightforward), or by calculating the Wronskian determinant.
\begin{align*} W(x) &= \det\begin{pmatrix}
y_1(x) & y_2(x)\\
y_1 '(x) & y_2 '(x)
\end{pmatrix} = \det\begin{pmatrix}
\frac{1}{x^2} & \frac{\log(x)}{x^2}\\
-\frac{2}{x^3} & \frac{1-2\log(x)}{x^3}
\end{pmatrix}\\
&= \frac{1 - 2\log(x) + 2\log(x)}{x^5} = \frac{1}{x^5}\neq 0 \end{align*}
Since the Wronskian determinant is not identically $0$, $y_1(x)$ and $y_2(x)$ form a fundamental system and are thus linearly independent.
