Let y = y 1 x and y = y 2 x be two distinct solutions of the differential equation d y d x = x + y , with y 1 0 = 0 and y 2 0 = 1 respectively. Then, the number of points of intersection of y = y 1 x and y = y 2 x is

Step-by-step solution

Given, d y d x = x + y ⇒ d y d x - y = x So, I F = e ∫ - 1 d x = e - x Now solution is given by y e - x = ∫ x e - x d x ⇒ y e - x = - x e - x - e - x + c ⇒ y = - x - 1 + c e x Given, y 1 0 = 0 ⇒ c = 1 ∴   y 1 = - x - 1 + e x             ⋯ 1 Also given y 2 0 = 1 ⇒ c = 2 ∴   y 2 = - x - 1 + 2 e x           ⋯ 2 Now from equation 1   &   2 we get, y 2 - y 1 = e x > 0 &#160