Let f(x)=x-1 and g(x)=e^x for x R . If d y d x = (e^ -2 x g(f(f(x)))- y x ), y(0)=0 , then y(1) is :-

Step-by-step solution

aligned & f(x)=x-1 & f(f(x))=f(x)-1=x-1-1=x-2 & g(f(f(x)))=e^ x-2 & d y d x =e^ -2 x e^ x-2 - 1 x y & d y d x + 1 x y=e^ x-2 x -2 which is L.D.E & I.F. =e^ d y x =e^ 2 x aligned Its solution is aligned & y e^ 2 x = e^ 2 x e^ x-2 x -2 d x+c & y e^ 2 x = e^ x-2 d x+c & y e^ 2 x =e^ x-2 +c aligned Given x =0, y =0 0= e ⁻²+ c ; c =- e ⁻² y e ^ 2 x = e ^ x -2 - e ⁻² when x =1, y e ^2= e ⁻¹- e ⁻² y= e⁻¹-e⁻² e^2 = 1 e - 1 e^2 e^2 = e^2-e e^5 = e-1 e^4 Option (1) is correct