Let y(x) be the solution of the differential equation x^2 d y d x +x y=x^2+y^2, x> 1 e satisfying y(1)=0 . Then the value of 2 (y(e))^2 y (e^2 ) is _____ .

Step-by-step solution

Put y = vx d y d x =v+x d v d x D.E. aligned & x ^2 (v+ x ~d v dx )+ x ^2 v= x ^2 (1+v^2 ) & v+x d v d x +v=1+v^2 & x d v d x =1+v^2-2 v & d v (v-1)^2 = d x x & - 1 v-1 = |x|+ C aligned x x - y = |x|+ C = x + C ( . Since . x > 1 e ) Given y(1)=0 C=1 So x x-y = (e x) Now y(e)= e 2 and y (e^2 )= 2 e^2 3 2(y(e))^2 y (e^2 ) = 2 e^2 4 2 e^2 3 = 3 4 =0.75