Let y=f(x) be the solution curve of the differential equation d y d x + y x = k + x (1 + k x)^2 , where k > 0 is a constant. If f(0) = 0 and f ( 3 ) = 3 14 , then f ( 4 ) is equal to :

Step-by-step solution

The given differential equation is linear with P(x) = x . Integrating Factor (IF) = e ^ x d x = e ^ ( x) = x . Multiplying the differential equation by the IF, we get: y x = k + x (1 + k x)^2 x d x Converting to cosine, we have: y x = k x + 1 ( x + k)^2 d x Observe that d d x ( x x + k ) = x( x + k) - x(- x) ( x + k)^2 = ^2 x + k x + ^2 x ( x + k)^2 = k x + 1 ( x + k)^2 . Thus, y x = x x + k + C . Given f(0) = 0 , we substitute x = 0 and y = 0 : 0 = 0 1 + k + C C = 0 . So, y = x x x + k . Given f ( 3 ) = 3 14 , we