If for the solution curve y=f(x) of the differential equation d y ~d x +( x) y= 2+ x (1+2 x)^2 , x ( - 2 , 2 ), f ( 3 )= 3 10 , then f ( 4 ) is equal to :

Step-by-step solution

aligned & If e ^ xdx = e ^ ( x ) = x & y x = 2+ x (1+2 x )^2 xdx & = 2 x +1 ( x +2)^2 dx Let x = 1- t ^2 1+ t ^2 & = 2 ( 1- t ^2 1+ t ^2 )+1 ( 1- t ^2 1+ t ^2 +2 )^2 2 dt aligned aligned & = 2-2 t^2+1+t^2 (1-t^2+2+2 t^2 )^2 2 dt & =2 3-t^2 (t^2+3 )^2 dt aligned Let t + 3 t = u aligned & (1- 3 t ^2 ) dt = du & =-2 du u ^2 aligned aligned & y ( x)= 2 u +c & y x= 2 t+ 3 t +c...(I) aligned At x = 3 , t = x 2 = 1 3 2 3 10 = 2 1 3 +3 3 + c aligned & 2. 3 10 = 2 3 10 + c C =0 & At x = 4 , t = x 2 = 2 -1 & y 2 = 2 2 -1+ 3