If a curve passes through the origin and the slope of the tangent to it at any point x , y is x 2 - 4 x + y + 8 x - 2 , then this curve also passes through the point:

Step-by-step solution

Given y 0 = 0 and d y d x = x - 2 2 + y + 4 x - 2 ⇒ d y d x - y x - 2 = x - 2 + 4 x - 2 ⇒   I . F . = e - ∫ 1 x - 2 d x = 1 x - 2 Solution of L.D.E. ⇒   y 1 x - 2 = ∫ 1 x - 2 x - 2 + 4 x - 2 · d x ⇒ y x - 2 = x - 4 x - 2 + C Now, at x = 0 , y = 0 ⇒ C = - 2 y = x x - 2 - 4 - 2 x - 2 ⇒ y = x 2 - 4 x This curve passes through 5 ,   5