Question

Solve the differential equation `"dy"/"dx" = 1 + "x"^2 +  "y"^2  +"x"^2"y"^2`, given that y = 1 when x = 0.

Answer 1

The given differential equation is:

`"dy"/"dx" = 1 + "x"^2 +  "y"^2  +"x"^2"y"^2`

`"dy"/"dx" (1 +"x"^2)(1+"y"^2)`

`⇒ "dy"/(1+"y"^2) = (1+"x"^2)"dx"`

Integrating both sides of this equation, we get:

`int "dy"/(1+"y"^2) = int (1+"x"^2)"dx"`

`⇒ tan^-1"y" = int"dx" + int"x"^2"dx"`

`⇒ tan^-1 "y" = "x"+"x"^3/3 +"C"`

It is given that y = 1 when x = 0.

`⇒ tan^-1 (1) = 0 + 0^3/3 + "C"`

`⇒ "C" = pi/4`

`⇒ tan^-1 "y" = "x" + "x"^3/3 + pi/4`

This is the required solution of the given differential equation.