use the substitution ( t=ln(x) ) to transform the given cauchy - euler equation to a differential equation…

Explanation:

Step1: Find the derivatives of (y) with respect to (x) in terms of (t)

Given (t = \ln(x)), then (x=e^{t}). By the chain - rule, (\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}). Since (\frac{dt}{dx}=\frac{1}{x}), we have (\frac{dy}{dx}=\frac{1}{x}\frac{dy}{dt}), so (x\frac{dy}{dx}=\frac{dy}{dt}). For the second - derivative, (\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(\frac{1}{x}\frac{dy}{dt})). Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = \frac{1}{x}) and (v=\frac{dy}{dt}), we get (\frac{d^{2}y}{dx^{2}}=-\frac{1}{x^{2}}\frac{dy}{dt}+\frac{1}{x^{2}}\frac{d^{2}y}{dt^{2}}). Then (x^{2}\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{dt^{2}}-\frac{dy}{dt}).

Step2: Substitute into the given Cauchy - Euler equation

The given equation is (x^{2}y''-3xy'+13y = 6 + 5x). Substitute (x^{2}y''=\frac{d^{2}y}{dt^{2}}-\frac{dy}{dt}), (xy'=\frac{dy}{dt}), and (x = e^{t}) into the equation: [ \begin{align*} \left(\frac{d^{2}y}{dt^{2}}-\frac{dy}{dt}\right)-3\frac{dy}{dt}+13y&=6 + 5e^{t}\ \frac{d^{2}y}{dt^{2}}-4\frac{dy}{dt}+13y&=6 + 5e^{t} \end{align*} ]

Answer:

(\frac{d^{2}y}{dt^{2}}-4\frac{dy}{dt}+13y = 6 + 5e^{t})