This problem is from the book "Introduction to Ordinary Differential Equations" by Shepley L. Ross. I am thinking I did the problem correctly because my answer matches the answer in the back of the book. What bothers me is the fact the second solution includes the first solution.
Problem:
Given that $y = x$ is a solution of
$$ x^2y'' – 4xy' + 4y = 0$$
find a linearly independent solution by reducing the order. Write the general solution.
Answer:
Let $f(x)$ represent the solution we have.
\begin{eqnarray*}
f(x) &=& x \\
y &=& f(x) v = xv \\
y' &=& x v' + v \\
y'' &=& x v'' + v' + v' = xv'' + 2v' \\
\end{eqnarray*}
\begin{eqnarray*}
x^2(xv'' + 2v') – 4x(x v' + v) + 4xv &=& 0 \\
x^3v'' + 2x^2v' – 4x^2v' &=& 0 \\
x^3v'' – 2x^2v' &=& 0 \\
\text{Let }w &=& \frac{dv}{dx} \\
x^3 w' – 2x^2 w &=& 0 \\
x w' – 2 w &=& 0 \\
x w' &=& 2w \\
x \frac{dw}{dx} &=& 2w \\
\frac{x}{dx} &=& \frac{2w}{dw} \\
\frac{x}{2dx} &=& \frac{w}{dw} \\
\frac{ 2dx}{x} &=& \frac{dw}{w} \\
2\ln{x} &=& \ln{w} + c_0 \\
x^2 &=& c_1 w \\
x^2 &=& c_1 \frac{dv}{dx} \\
\frac{dv}{dx} &=& c_2 x^2 \\
v &=& c_3 x^3 + c_4 \\
\frac{y}{x} &=& c_3 x^3 + c_4 \\
y &=& c_3 x^4 + c_4 x \\
\end{eqnarray*}
Hence the general solution is:
$$ y = C_0x^4 + C_1x $$
Best Answer
According to your question $ y = x $ is a solution of the given differential equation and after some procedure you get another solution of that equation as $ y = c_{3} x^{4} + c_{4} x $.
Now this second one is not the general solution but a solution of the given differential equation (although, in this particular differential equation it happens. But if you consider another differential equation (see Example $4.16$, "Introduction to Ordinary Differential Equations" by Shepley L. Ross, $3$rd Edition or, Example $2.10$ or Example $2.11$, "Differential Equations: Theory,Technique and Practice with Boundary Value Problems" by Simmons, George F. and Steven G. Krantz ) your concept will be more clear about the fact).
Now linear combination of these two solution is also a solution of the differential equation and so the general solution is $ y = C_{0} x^{4} + C_{1} x $. In this particular example both $x$ and $x^{4}$ satisfy the given equation and so both are solutions of the given differential equation.