JEE Main201815 Apr 2018Morning ShiftMathematicsDifferential EquationsActual
Let y=y(x) be the solution of the differential equation d y d x +2 y=f(x) , where f(x)= array lc 1, & x [0,1] 0, & otherwise array . If y(0)=0 , then y ( 3 2 ) is
Options
- Ae^2-1 2 e^3
- Be^2-1 e^3
- C1 2 e
- De^2+1 2 e^4
Correct answer
A. e^2-1 2 e^3
Step-by-step solution
When x [0,1] , then d y d x +2 y=1 aligned & y = 1 2 +C₁ e^ -2 x & y(0)=0 y(x)= 1 2 - 1 2 e^ -2 x & Here, y(1)= 1 2 - 1 2 e⁻²= e^2-1 2 e^2 aligned When x [0,1] , than d y d x +2 y=0 gathered y=c₂ e^ -2 x y(1)= e^2-1 2 e^2 e^2-1 2 =c^2 e⁻² C₂= e^2-1 2 y(x) ( e^2-1 2 ) e^ -2 x y ( 3 2 )= e^2-1 2 e^3 gathered