KCET2022MathematicsDifferential Equations
If y(x) is the solution of differential equation x x d y d x +y=2 x x, y(e) is equal to
Options
- Ae
- B0
- C2
- D2 e
Correct answer
C. 2
Step-by-step solution
x x d y d x +y=2 x x d y d x + y x x =2 aligned & IF =e^ 1 x x d x =e^ ( x) = x & y( x)= (2)( x) d x+C & y x=2 x d x+C & y x=2[x x-x]+C aligned Putting x=1 , we get y 0=2[1 0-1]+C C=2 Putting C=2 in Eq. (i), we get y x=2[x x-x]+2 At aligned x & =e y 1=2(e 1-e)+2 y & =0+2=2 y(e) & =2 aligned