Question

Solve the following Linear differential equation:

`(1 + x + xy^2) ("d"y)/("d"x) + (y + y^3)` = 0

Answer 1

`(1 + x + xy^2) ("d"y)/("d"x) = - (y + y^3)`

`(1 + x + xy^2) = - y(y^2 + 1) ("d"x)/("d"y)`

`y(y^2 + 1) ("d"x)/("d"y) + 1 + x(y^2 + 1)` = 0

Divided by `y(y^2 + 1)`,

`("d"x)/("d"y) + 1/(y(y^2 + 1)) + (x(y^2 + 1))/(y(y^2 + 1))` = 0

`("d"x)/("d"y) + x/y = - 1/(y(y^2 + 1))`

This is the form of `("d"x)/("d"y) + "P"x` = Q

Where P = `1/y` and Q = `(-1)/(y(1 + y^2))`

I.F = `"e"^(int "Pd"y)`

= `"e"^(int 1/y "d"y)`

= `"e"^(log y)`

= y

So, the solution of the equation is given by

`x xx "I.F" = int ("Q" xx "I.F")  "d"y + "c"`

`x xx y = int (-1)/(y(1 + y^2)) xx y xx "d"y + "c"`

xy = `int (-1)/(1 + y^2)  "d"y + "c"`

= `- int 1/(1 + y^2)  "d"y + "c"`

xy = `- tan^-1 "y" + "c"`

xy + `tan^-1 "y" + "c"`

Which is the required solution.