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#1
Given that y=arsinhx, show that

(1+x^2).d3y/dx3+3x.d^2y/dx^2=dy/dx=0

And also given that y=e^(arsinhx), show that

(1+x^2).d2y/dx2+x.dy/dx-y=0

I never get 0!!!!!

Also, what is the curve of a y=1/x graph calledd?
0
14 years ago
#2
rectangular hyperbola I think is the name
0
14 years ago
#3
(Original post by lgs98jonee)
Given that y=arsinhx, show that

(1+x^2).d3y/dx3+3x.d^2y/dx^2=dy/dx=0

And also given that y=e^(arsinhx), show that

(1+x^2).d2y/dx2+x.dy/dx-y=0

I never get 0!!!!!

Also, what is the curve of a y=1/x graph calledd?

y= arsinhx
y' = 1/rt(1 + x^2)
y'' = -x/(1 + x^2)^(3/2)
y''' = 3x^2/(1+x^2)^(5/2) - 1/(1+x^2)^(3/2)

so,

(1+x^2)y''' = (2x^2 - 1)/(1 + x^2)^(3/2)

3xy'' = -3x^2/(1+x^2)^(3/2)

y'' = 1/rt(1 + x^2)

=> (1+x^2)y''' + 3xy'' + y'' = (2x^2 - 1)/(1 + x^2)^(3/2) -3x^2/(1+x^2)^(3/2) + 1/rt(1 + x^2)

=> -1/rt(1 + x^2) + 1/rt(1 + x^2)
=> 0

================================ =================

y=e^(arsinhx) = x + rt(1+x^2)

y' = 1+ x/rt(1+x^2)

y'' = -x^2/(1+x^2)^(3/2) + 1/rt(1+x^2)

So,

(1+x^2)y'' + xy' - y = 1/rt(1+x^2) + x.(x+rt(1+x^2))/rt(1+x^2) - x - rt(1+x^2)

=> (1+x.rt(1+x^2) + x^2)/rt(1+x^2) - x - rt(1+x^2)
=> 0

================================ =================

y=1/x

is called a hyperbola.

================================ =================

Galois.
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