If the solution curve of the differential equation d y d x = x + y - 2 x - y passes through the point 2 , 1 and k + 1 , 2 , k > 0 , then

Step-by-step solution

Given, d y d x = x + y - 2 x - y = x - 1 + y - 1 x - 1 - y - 1 Now let x - 1 = X , y - 1 = Y So, d y d x = X + Y X - Y   . . . . . . . . 1 Now let Y = V X   d Y d X = V + X d V d X Putting the value in equation 1 we get, V + X d V d X = 1 + V 1 - V ⇒ X d V d X = V 2 + 1 1 - V ⇒ ∫ 1 - V 1 + V 2 d V = ∫ d X X ⇒ ∫ d V 1 + V 2 - 1 2 ∫ 2 V d V 1 + V 2 = ∫ d X X ⇒ tan - 1 V - 1 2 ln 1 + V 2 = ln X + c ⇒ tan - 1 Y X - 1 2 ln 1 + Y 2 X 2 = ln X + c &#865