# A Second Order Ordinary Differential equation with one known solution

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.
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: $$y = C_0x^4 + C_1x$$
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.