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maksim [4K]
3 years ago
10

Only answer if your 100% sure What equation is shown in the graph?

Mathematics
2 answers:
elena-s [515]3 years ago
6 0
(0,2) and (1,1)
slope = (2 - 1)/(0 -1) = -1
y intercept b = 2
equation
y = -x + 2

answer
equation
y = -x + 2
dedylja [7]3 years ago
6 0
Slope is -1 because from (0,2) to (1,1), that's m = slope = ( rise ) / ( run ) = ( 1 - 2 ) / (1 - 0) = -1 / 1 = -1

And y-intercept is y=2 at x=0, so b = 2

Putting it together:

y = mx+b

y = (-1)x + (2)

Final answer: y = -x + 2
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\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \  }

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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

8 0
4 years ago
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