Let y=y(x) be the solution of the differential equation (x^2+1 ) y^ -2 x y= (x^4+2 x^2+1 ) x , y(0)=1 . Then _ -3 ^3 y(x) d x is :

Step-by-step solution

aligned & (x^2+1 ) d y d x -2 x y= (x^4+2 x^2+1 ) x & d y d x - ( 2 x x^2+1 ) y= (x^2+1 )^2 x cx ^2+1 = ( x ^2+1 ) x & (Linear D.E) & P= -2 x x^2+1 , Q= (x^2+1 ) x & I.F =e^ P d x =e^ -2 x x^2+1 d x = 1 x^2+1 & y 1 x^2+1 = (x^2+1 ) x 1 x^2+1 d x & y x^2+1 = x+c y =1 c=1 & y= (x^2+1 )( x+1) & _ -3 ^3 y d x= _ -3 ^3 (x^2+1 )( x+1) & d x= _ -3 ^3 x^2 x+x^2 x+1 d x & _ -3 ^3 x^2 x d x+ _ -3 ^3 x^2 d x+ _ -3 ^3 x d x+ _ -3 ^3 1 d x & =0+18+0+6=24 aligned