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

Sum

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

#### Solution

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.

Concept: Methods of Solving First Order, First Degree Differential Equations - Differential Equations with Variables Separable Method
Is there an error in this question or solution?