Given d y d x +2 y x= x and y=0 for x= 3 . If maximum value of y is 1 k , then ' k ' is equal to

Step-by-step solution

d y d x +2 y x= x The given differential equation is of the form d y d x +P y=Q By comparing, we get P=2 x and Q= x I.F. =e^ 2 d x =e^ 2 x =e^ ^2 x = ^2 x Then, solution of the differential equation aligned & y ^2 x= x ^2 x d x+c & y ^2 x= x ^2 x d x+c & y ^2 x= x x d x+c & y ^2 x= x+c & y= 1 x + c ^2 x = x+ ( ^2 x ) c at & x= 3 , y=0 & 0= 3 + ( ^2 3 c ) & 0= 1 2 + c 4 c=-2 & y= x-2 ^2 x aligned y 2 =- ^2 x+ 1 2 x=- ( ^2 x- 1 2 x ) y 2 =- [ ( x- 1 4 )^2- ( 1 4 )^2 ]=- [ ( x- 1 4 )^2 ]+ 1 16 So, maximum value of y 2