A curve y=y(x) passes through the point ( , 0) . If the y -intercept of the tangent line to the curve at any point P(x, y) is equal to x^3 x , then the value of y ( 2 ) is

Step-by-step solution

The equation of the tangent line to the curve at a point (x, y) is given by: Y - y = dy dx (X - x) To find the y -intercept, we set X = 0 : Y = y - x dy dx According to the given condition: y - x dy dx = x^3 x x dy dx - y = -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 . The integrating factor (I.F.) is: I.F. = e^ - 1 x , dx = e^ - x = 1 x Multiplying by the I.F., the solution is given by: y 1 x = (-x^2 x ) ( 1 x ) dx y x = - x x , dx Usin