If the general solution of the differential equation ^2 x d y d x +y= x is y= x-1+C e^ - x satisfies y ( 4 )=1 , then C=

Step-by-step solution

Given, ^2 x d y d x +y= x d y d x +y ^2 x= x ^2 x ...(i) Here, p= ^2 x p d p= ^2 x d x= xI F=e^ x Multiplying Eq. (i) by I F , we get e^ x d y d x +e^ x y ^2 x=e^ x x ^2 x Integrating both sides, we get y e^ x = e^ x x ^2 x d x On putting x=t , ^2 x d x=d t y e^t= t e^t d t=e^t(t-1)+C y e^ x = x-1+C e^ - x If y ( 4 )=1 1=1-1+C e⁻¹ 1=C e⁻¹ C=e