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I am asked to find the general solution of the differential equation:

\begin{equation*} y^{\prime\prime\prime}(x)-\frac{2}{x}y^{\prime\prime}(x)+y^{\prime}(x)-\frac{2}{x}y(x) = 0 \end{equation*} Given two known solutions $y_{1} = \cos(x)$ and $y_{2} = \sin(x)$.

My understanding is that the general solution will be a linear combination of independent solutions with the form:

\begin{equation*} y(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x) + c_{3}y_{3}(x) \end{equation*}

Using the lecture notes from my ODE class as guidance, it appears that I'm suppose to use the Wronskian to find the general solution. Thus, I compute the Wronskian W(x):

\begin{alignat*}{2} W(x) &= \left| \begin{array}{ccc} \cos(x) & \sin(x) & y_{3} \\ -\sin(x) & \cos(x) & y_{3}^{\prime} \\ -\cos(x) & -\sin(x) & y_{3}^{\prime\prime} \end{array} \right| \\ &= \left| \begin{array}{ccc} \cos(x) & \sin(x) & y_{3} \\ -\sin(x) & \cos(x) & y_{3}^{\prime} \\ 0 & 0 & y_{3}^{\prime\prime}+y_{3} \end{array} \right| \\ &\ne 0 \end{alignat*}

and find that

\begin{alignat*}{2} (y_{3}^{\prime\prime}+y_{3})(\cos^{2}(x)+\sin^{2}(x)) &\ne 0 &&\Rightarrow \\ (y_{3}^{\prime\prime}+y_{3})\cdot 1 &\ne 0 \end{alignat*}

However, from here, I'm not sure what to do next since I only know that $W(x)$ should be nonzero. I also notice that $y\equiv 0$ also looks like a solution, but I'm not sure if that can be used in some way.

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Using Liouville's Formula

\begin{alignat*}{2} W(x) &= W(x_{0})e^{-\int_{x_{0}}^{x}a_{1}(t)dt} \\ &= W(x_{0})e^{2\int_{x_{0}}^{x}\frac{1}{t}dt} \\ &= W(x_{0})e^{2\ln(x)-2\ln(x_{0})} \\ &= W(x_{0})\left(x^{2}\cdot x_{0}^{-2}\right) \\ &= cx^{2} &&\Rightarrow \\ y^{\prime\prime}_{3} + y &= cx^{2} \end{alignat*}

Then combining a homogeneous and particular solution gives \begin{equation*} y(x) = cx^{2}-2c + k_{1}\cos(x) + k_{2}\sin(x) \end{equation*}

and subtracting out $y_{1}$ and $y_{2}$ gives \begin{alignat*}{2} y_{3}(x) &= cx^{2}-2c &&\Rightarrow \\ c_{3}y_{3}(x) &= c_{3}(x^{2}-2). \end{alignat*}

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