JEE Main20199 Apr 2019Morning ShiftMathematicsDifferential EquationsActual
The solution of the differential equation x d y d x + 2 y = x 2 , ( x ≠ 0 ) with y 1 = 1 , is
Options
- Ay = x 3 5 + 1 5 x 2
- By = 3 4 x 2 + 1 4 x 2
- Cy = x 2 4 + 3 4 x 2
- Dy = 4 5 x 3 + 1 5 x 2
Correct answer
C. y = x 2 4 + 3 4 x 2
Step-by-step solution
Given, differential equation is d y d x + 2 x y = x This is a linear differential equation of type d y d x + P y = Q , where P   &   Q are the functions of x or constants. Thus, P = 2 x   &   Q = x The integrating factor I . F . = e ∫ P d x = e ∫ 2 x   d x = e 2 ln x = e ln x 2 = x 2 . The solution of the linear differential equation is y × I . F . = ∫ Q × I . F . d x + C So, the solution of the given differential equation is y x 2 = ∫ x · x