Answer:
[tex]\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }[/tex]
Step-by-step explanation:
Hello,
let's follow the advise and proceed with the substitution
first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t
[tex]x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)<=>y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})[/tex]
Now we can substitute in the equation
[tex]x^2y''(x)+9xy'(x)-20y(x)=0\\<=> e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\<=> \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\<=> \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\[/tex]
so the new equation is
[tex]y''(t)+ 8y'(t)-20y(t)=0[/tex]
the auxiliary equation is
[tex]x^2+8x-20=0\\<=> x^2-2x+10x-20=0\\<=>x(x-2)+10(x-2)=0\\<=>(x+10)(x-2)=0\\<=> x=-10\text{ or }x=2[/tex]
so the solutions of the new equation are
[tex]y(t)=ae^{2t}+be^{-10t}[/tex]
with a and b real
as
[tex]x(t)=e^t\\<=> t(x)=ln(x)[/tex]
[tex]y(x)=ae^{2ln(x)}+be^{-10ln(x)}=ax^2+bx^{-10}[/tex]
hope this helps
do not hesitate if you have any questions
