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solve the differential equation : `[ sqrt(x+y)+sqrt(x-y) ]dx + [sqrt(x+y)-sqrt(x-y) ]dy...
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`[sqrt(x+y)+sqrt(x-y)]dx + [sqrt(x+y)-sqrt(x-y)]dy = 0`
We can easily do this using a substitution.
`sqrt(x+y) = u`
`sqrt(x-y) = v`
`x+y = u^2 and x-y = v^2`
`2x = u^2+v^2`
`2y = u^2-v^2`
Differentiating above equations would give,
`2dx = 2udu+2vdv ---gt dx = udu+vdv`
`2dy = 2udu-2vdv ---gt dy = udu-vdv`
Therefore the equation becomes,
`(u+v)(udu+vdv)+(u-v)(udu-vdv) = 0`
`u^2du+uvdv+uvdu+v^2dv+u^2du-uvdv-uvdu+v^2dv = 0`
`2u^2du+2v^2 dv= 0`
Now we can separate the variables,
`u^3/3 = -v^3/3 + c`
`u^3/3+v^3/3 = c`
`u^3+v^3 = 3c = k`
Therefore the solution is,
`(sqrt(x+y))^3 + (sqrt(x-y))^3 = k`
`(x+y)^(3/2) + (x-y)^(3/2) = k`
Where k is an arbitrary constant.
Posted by thilina-g on June 14, 2012 at 6:55 PM (Answer #1)
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