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x`sqrt(1+y^2)dx-ysqrt(1+x^2)dy=0` solve using variable separable.

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dylzzz | Student, Undergraduate | (Level 1) Honors

Posted July 1, 2013 at 2:41 PM via web

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x`sqrt(1+y^2)dx-ysqrt(1+x^2)dy=0`

solve using variable separable.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 1, 2013 at 3:17 PM (Answer #1)

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You need to start separate the variables, hence, you need to move the negative member to the right side, such that:

`xsqrt(1 + y^2)dx = ysqrt(1+x^2)dy`

You need to divide by `sqrt(1 + y^2)` both sides, such that:

`xdx = (ysqrt(1+x^2)dy)/(sqrt(1 + y^2))`

You need to divide by `sqrt(1 + x^2)` both sides, such that:

`(xdx)/(sqrt(1 + x^2)) = (ydy)/(sqrt(1 + y^2))`

Integrating both sides yields:

`int (xdx)/(sqrt(1 + x^2)) = int (ydy)/(sqrt(1 + y^2))`

You should come up with the following substitutions, such that:

`1 + x^2 = t => 2xdx = dt => xdx = (dt)/2`

`1 + y^2 = u => 2ydy = du => ydy = (du)/2`

Changing the variables, yields:

`int ((dt)/2)/(sqrt t) = int ((du)/2)/(sqrt u)`

`(1/2)int (dt)/(sqrt t) = (1/2) int (du)/(sqrt u)`

`(sqrt t)/(1/2) = (sqrt u)/(1/2) => sqrt t = sqrt u => t = u`

Replacing back `1 + x^2` for t and `1 + y^2` for u yields:

`1 + x^2 = 1 + y^2 => x^2 = y^2 => y = sqrt(x^2) + c => y = |x| + c`

Hence, evaluating the solution to the given differential equation, using separation of variables, yields the general solution `y = |x| + c.`

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