(2x/y - y/x^2+y^2)dx + (x/x^2+y^2 - x^2/y^2)dy =0 its differential equation . 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should write the equation such that:

`2x/y - y/(x^2+y^2) + (x/(x^2+y^2) - x^2/y^2) (dy)/(dx) = 0`

You should come up with the following notations such that:

`F(x,y) = 2x/y - y/(x^2+y^2)`

`G(x,y) = x/(x^2+y^2) - x^2/y^2`

You need to check if the equation is exact such that:

`(delF(x,y))/(del y) = -2x/(y^2) - (x^2+y^2 - 2y^2)/((x^2+y^2)^2)`

`(delF(x,y))/(del y) = -2x/(y^2) - 1/(x^2+y^2) + (2y^2)/((x^2+y^2)^2)`

`(delG(x,y))/(del x) = (x^2+y^2-2x^2)/((x^2+y^2)^2) - 2x/y^2`

`(delG(x,y))/(del x) = 1/(x^2+y^2) - (2x^2)/((x^2+y^2)^2) - 2x/y^2`

Notice that `(delF(x,y))/(del y) = (delG(x,y))/(del x),`  hence, the equation is exact and you need to define f(x,y) such that:

`f(x,y) = c`

`f(x,y) = int (2x/y - y/(x^2+y^2)) dx = (1/y)x^2 - y int 1/(x^2+y^2)) dx `

`int (2x/y - y/(x^2+y^2)) dx = (1/y)x^2 - y*(1/y)*arctan (x/y) + c(y)`

`int (2x/y - y/(x^2+y^2)) dx = (1/y)x^2 - arctan (x/y) + c(y)`

You need to differentiate `f(x,y)`  with respect to y to determine c(y) such that:

`(del f(x,y))/(del y) = -x^2/y^2 - (x/y^2)/(1 + (x/y)^2) + (dc(y))/(dy)`

`(del f(x,y))/(del y) = -x^2/y^2 - x/(x^2+y^2) + (dc(y))/(dy)`

Considering `(dc(y))/(dy) = 0`  yields:

`int (dc(y))/(dy) dy = 0 => c(y) = 0 => f(x,y) = c = (1/y)x^2 - arctan (x/y)`

Hence, evaluating the solution to the first order nonlinear differential equation yields `c = (x^2/y) - arctan (x/y).`

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