The slope of the tangent to a curve y = y(x) at any point (x, y) is given by x^2 x - y x for x > 0 . If the curve passes through the point ( , 0) , then the value of y ( 2 ) is equal to

Step-by-step solution

The given geometric condition translates to the differential equation: dy dx = x^2 x - y x Rearranging this into the standard linear differential equation form: dy dx + 1 x y = x x The integrating factor (I.F.) is: I.F. = e^ 1 x dx = e^ x = x Multiplying the equation by the I.F. and integrating: y x = x^2 x , dx Applying integration by parts twice on the right side: x^2 x , dx = x^2(- x) - 2x(- x) , dx = -x^2 x + 2 x x , dx = -x^2 x + 2 (x x - x , dx ) = -x^2 x + 2x x + 2 x + C So, the general solution is: xy = -x^