NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The solution of the differential equation d y d x = y 2 + x ln x 2 x y is (where, c is the constant of integration)
Options
- A2 x 2 = y ln x 2 + 2 c y
- B2 y 2 = x ln x 2 + 2 c x
- Cx 2 = y ln x 2 + c
- D2 y 2 = x y ln x 2 + c x
Correct answer
B. 2 y 2 = x ln x 2 + 2 c x
Step-by-step solution
2 x y d y d x = y 2 + x ln ⁡ x 2 y d y d x - y 2 x = ln ⁡ x Put, y 2 = t ⇒ 2 y d y d x = d t d x d t d x - t x = ln ⁡ x I.F. = e ∫ - 1 x d x = e - l n x = 1 x The solution is t x = ∫ ln ⁡ x x d x t x = ln ⁡ x 2 2 + c y 2 x = ln ⁡ x 2 2 + c 2 y 2 = x ln ⁡ x 2 + 2 c x