how to solve (y^2 cos x-y)dx + ((x+y^2)dy  resolve by integral factor. please I´need help

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use the following notation `F(x,y) = y^2 cos x - y`  and `G(x,y) = x + y^2` .

Differentiating the function F with respect to y and the function G with respect to x yields:

`(delF)/(del y) = 2y cos x - 1`

`(delG)/(del x)= 1`

Notice that `(delF)/(del y) != (delG)/(del x), ` hence the given equation is not an exact equation.

You need to find n integrating factor `mu(y)`  such that:

`F(x,y)mu(y) + mu(y)*(dy)/(dx)G(x,y) = 0`

`(y^2 cos x - y)(dmu(y))/(dy) + mu(y)(2y cos x - 1) = mu(y) ` `(y^2 cos x - y)(dmu(y))/(dy) = mu(y) - mu(y)(2y cos x - 1)`

`(y^2 cos x - y)(dmu(y))/(dy) = mu(y)(2 - 2y cos x)`

`(y^2 cos x - y)(dmu(y))/(dy) = 2mu(y)(1 - ycos x)`

`y(ycos x - 1)(dmu(y))/(dy) = 2mu(y)(1 - ycos x)`

`(dmu(y))/(dy) /(mu(y)) = (2(1 - ycos x))/(y(ycos x - 1))`

`(dmu(y))/(dy) /(mu(y)) = -2/y`

Integrating both sides yields:

`int (dmu(y))/(dy) /(mu(y)) = int -2/y dy`

`ln(mu(y)) = -2ln y + c => mu(y) = y^(-2) => mu(y) = 1/(y^2)`

You need to multiply your equation by integrating factor mu(y) such that:

`-1/y + (x/(y^2) + 1)(dy)/(dx) + cos x = 0`

Selecting `F(x,y) = cos x - 1/y`  and `G(x,y) = x/(y^2)+1` , notice that `(delF)/(del y) = (delG)/(del x) = 1/(y^2)` , hence, this is an exact equation, having as solution the function `f(x,y) = int (cos x - 1/y)dx`  such that:

`f(x,y) = int (cos x - 1/y)dx = sin x - x/y + c(y)`

Since `(dc(y))/(dy) = 1 =>f(x,y) = sin x - x/y + y`

Substituting c for f(x,y) yields:

`sin x - x/y + y = c => y sin x - x + y^2 = cy => y^2 - y(sin x - c) - x = 0`

You should use quadratic formula such that:

`y_(1,2) = (sin x - c +- sqrt(sin^2 x - 2c sin x + c^2 + 4x))/2`

Hence, evaluating the solutions to the first order nonlineary differential equation yields `y_(1,2) = (sin x - c +- sqrt(sin^2 x - 2c sin x + c^2 + 4x))/2.`

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