Let a curve x = x(y) pass through the origin and satisfy the differential equation (1+y^2)dx + (x - e^ (y) )dy = 0 . Then the value of x(1) is equal to:

Step-by-step solution

The given differential equation is (1+y^2)dx + (x - e^ (y) )dy = 0 . Rearranging the terms, we get: dx dy + x 1+y^2 = e^ (y) 1+y^2 This is a linear differential equation of the form dx dy + P(y)x = Q(y) , where P(y) = 1 1+y^2 . The integrating factor (I.F.) is: I.F. = e^ 1 1+y^2 dy = e^ (y) Multiplying the equation by the I.F. and integrating, we get: x e^ (y) = e^ 2 (y) 1+y^2 dy Let t = (y) , then dt = 1 1+y^2 dy . The integral becomes: e^ 2t dt = 1 2 e^ 2t + C = 1 2 e^ 2 (y) + C So, the general solution is: x e^