A curve y = f(x) in the first quadrant passes through the point (1, 1) . If the y -intercept of the tangent line to the curve at any point P(x, y) is given by xy - x^4 , then the value of f(3) is equal to :

Step-by-step solution

The equation of the tangent line at any point (x, y) on the curve is 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 = xy - x^4 . Rearranging this, we get x dy dx + (x - 1)y = x^4 , which can be rewritten as dy dx + (1 - 1 x )y = x^3 . This is a linear differential equation with integrating factor IF = e^ (1 - 1 x ) dx = e^ x - x = e^x x . Multiplying both sides by the integrating factor and integrating: d (y e^x x ) = x^3