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xdx-`sqrt(a^2+x^2)` dy=0Using seperation of variable.

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dylzzz | Student, Undergraduate | Honors

Posted July 18, 2013 at 8:25 PM via web

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xdx-`sqrt(a^2+x^2)` dy=0
Using seperation of variable.

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llltkl | College Teacher | Valedictorian

Posted July 19, 2013 at 1:16 AM (Answer #1)

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Given `xdx-sqrt(a^2+x^2) dy=0`

Separating variables gives us:

`dy=(xdx)/sqrt(a^2+x^2)`

Upon integrating :

`int dy= int (xdx)/sqrt(a^2+x^2)`

let `u=a^2+x^2`   then `du=2xdx rArr (du)/2=xdx`

So, `int dy= 1/2 int (du)/sqrtu`

`rArr y=1/2*2*u^(1/2)+c`

`rArr y= sqrt(a^2+x^2) + c` `=>` answer.

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