Question

Find the equation of a curve passing through origin and satisfying the differential equation `(1 + x^2) "dy"/"dx" + 2xy` = 4x² 

Answer 1

Given equation is `(1 + x^2) "dy"/"dx" + 2xy` = 4x² 

⇒ `"dy"/"dx" + (2x)/(1 + x^2) * y = (4x^2)/(1 + x^2)`

Here, P = `(2x)/(1 + x^2)` and Q = `(4x^2)/(1 + x^2)`

Integrating factor I.F. = `"e"^(int Pdx)`

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

= `"e"^(log(1 + x^2)`

= 1 + x²

∴ Solution is `y xx "I"."F". = int "Q" xx "I"."F".  "d"x + "c"`

⇒ `y(1 + x^2) = int (4x^2)/(1 + x^2) xx (1 + x^2) "d"x + "c"`

⇒ `y(1 + x^2) = int 4x^2 "d"x + "c"`

⇒ `y(1 + x^2) = 4/3 x^3 + "c"`  ......(i)

Since the curve is passing through origin i.e., (0, 0)

∴ Put y = 0 and x = 0 in equation (i)

0(1 + 0) = `4/3(0)^3 + "c"`

⇒ C = 0

∴ Equation is `y(1 + x^2) = 4/3 x^3`

⇒ y = `(4x^3)/(3(1 + x^2))`

Hence, the required solution is y =  `(4x^3)/(3(1 + x^2))`.