Let y=y(x) be the solution of the differential equation x ~d y ~d x +y=x x,(x 1) If 2(y(2))= 4-1 then the value of y( e ) is

Step-by-step solution

aligned & x ~d y ~d x +y=x x & ~d y ~d x + 1 x y= x aligned Here, P (x)= 1 x , Q (x)= x array ll & I.F. = e ^ P (x) dx = e ^ 1 x dx = e ^ x =x & y( I.F. )= Q ( I.F. ) d x+ c & x y= x x ~d x+ c & x y= x^2 2 x- ( x^2 2 1 x ~d x )+ c & x y= x^2 2 x- 1 2 x ~d x+ c & x y= x^2 2 x- 1 4 x^2+ c ...(i) array Given that 2(y(2))= 4-1 y(2)= 2- 1 2 From equation (i), we get array ll & 2 ( 2- 1 2 )= (2)^2 2 2- 1 4 (2)^2+ c & 2 2-1=2 2-1+c & c =0...(ii) array From (i) and (ii), we get y( e )= e 2 - e 4 = e 4