JEE Main202312 Apr 2023Morning ShiftMathematicsDifferential EquationsActual
Let y = y x , y > 0 , be a solution curve of the differential equation 1 + x 2 d y = y x - y d x . If y 0 = 1 and y 2 2 = β , then
Options
- Ae 3 β - 1 = e 3 + 2 2
- Be 3 β - 1 = e 5 + 2
- Ce β - 1 = e - 2 3 + 2 2
- De β - 1 = e - 2 5 + 2
Correct answer
A. e 3 β - 1 = e 3 + 2 2
Step-by-step solution
Given, 1 + x 2 d y = y x - y d x ⇒ d y d x = x 1 + x 2 y - y 2 1 + x 2 ⇒ - 1 y 2 d y d x + x 1 + x 2 · 1 y = 1 1 + x 2 Now let 1 y = t , we get, - 1 y 2 y ' = d t d x So, the equation becomes, d t d x + x 1 + x 2 t = 1 1 + x 2 Now finding, IF = e ∫ x 1 + x 2 d x = 1 + x 2 So, solution of the differential equation is given by, t 1 + x 2 = ∫ 1 + x 2 1 + x 2 d x ⇒ 1 y 1 + x 2 = ∫ 1 1 + x 2 d x ⇒ 1 y 1 + x 2 = ln x + x 2 + 1 + C ∵   y 0 = 1 ⇒ C = 1