NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The solution of the differential equation x d y - y d x + 3 x 2 y 2 e x 3 d x = 0 is (where, c is an arbitrary constant)
Options
- Ax = 2 y e x + c
- Bx = y e x 3 + c y
- Cx = y 2 e x 3 + c
- Dx y = e x 3 + c
Correct answer
B. x = y e x 3 + c y
Step-by-step solution
Given equation is y d x - x d y y 2 = 3 x 2 e x 3 d x or d x y = d e x 3 ⇒ ∫ d x y = ∫ d e x 3 ⇒ x y = e x 3 + c or x = y e x 3 + c y