Use the substitution of x=e^{t} to transform the given Cauchy-Euler differential equation to a differential equation with constant coefficients then solve that by any method. (finding the characterristic (auxiliary) equation or variation of parameters). Do not forget to re-subtitute for x after you solve the equation.x^{2}y^{''}+9xy^{'}-20y=0

By the chain rule,[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}\implies\dfrac{\mathrm dy}{\mathrm dt}=x\dfrac{\mathrm dy}{\mathrm dx}[/tex]which follows from [tex]x=e^t\implies t=\ln x\implies\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x[/tex].[tex]\dfrac{\mathrm dy}{\mathrm dt}[/tex] is then a function of [tex]x[/tex]; denote this function by [tex]f(x)[/tex]. Then by the product rule,[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac1x\dfrac{\mathrm df}{\mathrm dx}[/tex]and by the chain rule,[tex]\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dt^2}[/tex]so that[tex]\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}=x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}[/tex]Then the ODE in terms of [tex]t[/tex] is[tex]\dfrac{\mathrm d^2y}{\mathrm dt^2}+8\dfrac{\mathrm dy}{\mathrm dt}-20y=0[/tex]The characteristic equation[tex]r^2+8r-20=(r+10)(r-2)=0[/tex]has two roots at [tex]r=-10[/tex] and [tex]r=2[/tex], so the characteristic solution is[tex]y_c(t)=C_1e^{-10t}+C_2e^{2t}[/tex]Solving in terms of [tex]x[/tex] gives[tex]y_c(x)=C_1e^{-10\ln x}+C_2e^{2\ln x}\implies\boxed{y_c(x)=C_1x^{-10}+C_2x^2}[/tex]