NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
Let y x is the solution of the differential equation x + 2 d y d x - x + 1 y = 2 . If y 0 = - 1 , then the value of y 2 is equal to
Options
- Ae 2 + 1 2
- Be 2 - 1 2
- C1 2 - e 2
- De 2
Correct answer
B. e 2 - 1 2
Step-by-step solution
Given differential equation is d y d x - x + 1 x + 2 y = 2 x + 2 I.F. = e - ∫ x + 1 x + 2 d x = e - x x + 2 So the solution of the equation is y x + 2 e - x = - 2 e - x + C Since y 0 = 1 , we have C = 4 So y 2 = e 2 - 1 2