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