Let y = y(x) be the solution curve of the differential equation x d y - y d x = x^2+y^2 d x , where x > 0 . If the curve intersects the line y = 3x 4 at the point where x = 2 , then the value of y when x = 3 is equal to

Step-by-step solution

Given differential equation is x d y - y d x = x^2+y^2 d x d y d x = y x + 1 + ( y x )^2 Let y = vx d y d x = v + x d v d x Substituting in the differential equation, we get: v + x d v d x = v + 1+v^2 x d v d x = 1+v^2 d v 1+v^2 = d x x Integrating both sides: d v 1+v^2 = d x x (v + 1+v^2 ) = x + C v + 1+v^2 = A x (where A = e^C ) Substituting v = y x : y x + 1 + y^2 x^2 = A x y + x^2+y^2 = A x^2 The curve intersects y = 3x 4 at x = 2 . So, y = 3(2) 4 = 3 2 . The point is (2, 3 2 ) . Substituting x = 2, y = 3 2 : 3