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.


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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.