Let x=x(y) be the solution of the differential equation y^2 ~d x+ (x- 1 y ) d y=0 . If x(1)=1 , then x ( 1 2 ) is :

Step-by-step solution

aligned & y^2 d x+ (x- 1 y ) d y=0 & y^2 d x= ( 1 y -x ) d y & y^2 d x d y = 1 y -x & d x d y + x y^2 = 1 y^3 & I.F. =e^ 1 y^2 d y =e^ -1 y aligned Solution is x e^ -1 y = e^ - 1 y 1 y^3 d y+C Let -1 y =t aligned & 1 y^2 d y=d t & x e^ - 1 y =- e^t t d t+C & x e^ - 1 y =-e^t(t-1)+C & x e^ - 1 y =-e^ -1 y ( -1 y -1 )+C aligned x(1)=1 e⁻¹=-e⁻¹(-2)+C C=-e⁻¹ x= 1 y +1-e^ -1+ 1 y x ( 1 2 )=3-e