What is 5(x-1)= 20(x-10)

Mathematics

2 answers:

Dmitry [639]4 years ago

7 0

X = 41/5 or 8 1/5 my guyyyyyyyy

valentinak56 [21]4 years ago

5 0

Answer:

I got an answer of X = 13.

You might be interested in

Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

Anika [276]

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

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