If y=y(x) is the solution of d y d x = x-y x 1+ x , y ( 2 )= ^2 8 , then y( )=

Step-by-step solution

d y d x = x-y x 1+ x d y d x = x 1+ x -y x 1+ x d y d x + x 1+ x y= x 1+ x ...(i) Eqn. (i) is a linear differential equation of first order and first degree. Its integrating factor is I . F .=e^ x 1+ x +x =e^ (1+ x) I.F =1+ x Now, the solution of eqn. (i) can be written as : y (1+ x)= x (1+ x) (1+ x) d x+C y(1+ x)= x^2 2 +C ...(ii) At x= 2 , y= ^2 8 Then from (ii), we get : ^2 8 (1+1)= ^2 4 2 +C C= ^2 8 Then eqn. (ii) becomes : y(1+ x)= x^2 2 + ^2 8 Now, at x= : y( ) (1+0)= ^2 2 + ^2 8 y( )= 5 ^2 8 .