JEE Main202127 Aug 2021Morning ShiftMathematicsDifferential EquationsActual
Let us consider a curve, y = f x passing through the point - 2 , 2 and the slope of the tangent to the curve at any point ( x , f ( x ) ) is given by f ( x ) + x f ' ( x ) = x 2 . Then
Options
- Ax 3 - 3 x f ( x ) - 4 = 0
- Bx 2 + 2 x f ( x ) - 12 = 0
- Cx 3 + x f ( x ) + 12 = 0
- Dx 2 + 2 x f ( x ) + 4 = 0
Correct answer
A. x 3 - 3 x f ( x ) - 4 = 0
Step-by-step solution
y + x d y d x = x 2 d y d x + y x = x I . F . = e ∫ 1 x d x = x Solution of differential equation, y · x = ∫ x · x   d x x y = x 3 3 + C Passes through - 2 ,   2 , So, - 12 = - 8 + 3 C C = - 4 3 ∴   3 x y = x 3 - 4 i.e 3 x · f x = x 3 - 4 x 3 - 3 x   f x - 4 = 0