Let a curve y = y(x) have the property that the y -intercept of the tangent to the curve at any point (x, y) is equal to -x^3 x . If the curve passes through the point ( , 0) , then the value of y ( 2 ) is equal to

Step-by-step solution

The equation of the tangent at (x, y) to the curve is given by Y - y = dy dx (X - x) . To find the y -intercept, we set X = 0 , which gives Y = y - x dy dx . According to the given condition, y - x dy dx = -x^3 x dy dx - 1 x y = x^2 x This is a linear differential equation of the form dy dx + P(x)y = Q(x) , where P(x) = - 1 x and Q(x) = x^2 x . Integrating Factor (I.F.) = e^ - 1 x dx = e^ - x = 1 x . The general solution is given by: y 1 x = (x^2 x 1 x ) dx + C y x = x x dx + C Using integration by parts: y x = x x