Let a curve y=y(x) satisfy the differential equation (1+x^2) dy dx = y(y-1) x with the initial condition y(0) = -1 . If the area of the triangle formed by the tangent to the curve at x = 3 and the coordinate axes is a 3 b , where a and b are coprime positive integers, then the value of a+b is equal to:

Step-by-step solution

Given differential equation: (1+x^2) dy dx = y(y-1) x Separating the variables: dy y(y-1) = x 1+x^2 dx Using partial fractions on the left side: ( 1 y-1 - 1 y ) dy = x 1+x^2 dx |y-1| - |y| = 1 2 (1+x^2) + C | y-1 y | = 1 2 (1+x^2) + C Using the initial condition y(0) = -1 : | -2 -1 | = 1 2 (1) + C C = 2 To find the y -coordinate at x = 3 : | y-1 y | = 1 2 (1+3) + 2 = 2 + 2 = 4 | y-1 y | = 4 Since y(0) = -1 , the curve lies in the region y 1 , so we take the positive sign: 1 - 1 y = 4 - 1 y = 3 y = - 1 3 To find the