Let a curve y = f(x) pass through the point (1, e) . If the y -intercept of the tangent to the curve at any point (x, y) is given by -x^3 e^x , then the value of f(2) is equal to

Step-by-step solution

The equation of the tangent to the curve y = f(x) at a point (x, y) is: Y - y = dy dx (X - x) To find the y -intercept, set X = 0 : Y = y - x dy dx According to the given condition: y - x dy dx = -x^3 e^x x dy dx - y = x^3 e^x dy dx - 1 x y = x^2 e^x This is a linear differential equation with P(x) = - 1 x and Q(x) = x^2 e^x . Integrating factor (IF) = e^ - 1 x dx = e^ - x = 1 x The general solution is: y ( 1 x ) = x^2 e^x ( 1 x ) dx + C y x = x e^x dx + C Using integration by parts: y x = x e^x - 1 e^x dx + C y x