Question

Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients. Solve the original equation by solving the new equation
x2y'' + 9xy' - 20y = 0

Answer 1

Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

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

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0$

so the new equation is

$y''(t)+ 8y'(t)-20y(t)=0$

the auxiliary equation is

$x^2+8x-20=0\\ x^2-2x+10x-20=0\\x(x-2)+10(x-2)=0\\(x+10)(x-2)=0\\ x=-10\text{ or }x=2$

so the solutions of the new equation are

$y(t)=ae^{2t}+be^{-10t}$

with a and b real

as

$x(t)=e^t\\ t(x)=ln(x)$

$y(x)=ae^{2ln(x)}+be^{-10ln(x)}=ax^2+bx^{-10}$

hope this helps

do not hesitate if you have any questions