# solve the differential equation : `[ sqrt(x+y)+sqrt(x-y) ]dx + [sqrt(x+y)-sqrt(x-y) ]dy = 0`

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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.

Let,

`sqrt(x+y) = u`

and

`sqrt(x-y) = v`

Then,

`x+y = u^2 and x-y = v^2`

This gives,

`2x = u^2+v^2`

and

`2y = u^2-v^2`

Differentiating above equations would give,

`2dx = 2udu+2vdv ---gt dx = udu+vdv`

and

`2dy = 2udu-2vdv ---gt dy = udu-vdv`

Therefore the equation becomes,

`(u+v)(udu+vdv)+(u-v)(udu-vdv) = 0`

Removing brackets,

`u^2du+uvdv+uvdu+v^2dv+u^2du-uvdv-uvdu+v^2dv = 0`

This gives,

`2u^2du+2v^2 dv= 0`

Now we can separate the variables,

`u^2du=-v^2dv`

Integrating,

`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.