Let x = x(y) be the solution of the differential equation dy dx = y^2 xy - 4 . If the solution curve passes through the point (5, 2) , then the minimum value of x(y) for y > 0 is equal to :

Step-by-step solution

The given differential equation can be rewritten by taking the reciprocal: dx dy = xy - 4 y^2 = x y - 4 y^2 Rearranging terms, we get a linear differential equation in x : dx dy - 1 y x = - 4 y^2 The integrating factor is IF = e^ - 1 y dy = e^ - y = 1 y . Multiplying the equation by the integrating factor and integrating: d (x 1 y ) = - 4 y^3 dy x y = -4 ( y⁻² -2 ) + C = 2 y^2 + C Thus, the general solution is x = 2 y + Cy . The curve passes through (5, 2) , so we substitute x = 5 and y = 2 : 5 = 2 2 + C(2) 5 = 1 +