Question

Solution of (dy/dx) = 1 + x + y2 + xy2, y(0) = 0 is:

(A) y = tan(c + x + x2)

(B) y = tan [x + (x2/2)]

(C) y2 = exp[x + (x2/2)]

(D) y2 = 1 + c ∙ exp[x + (x2/2)]

Answer 1

The correct option (B) y = tan [x + (x2/2)]

Explanation:

(dy/dx) = 1 + x + y2 + xy2 (dy/dx) = (1 + x) (1 + y2)

∴ [(dy)/(1 + y2)] = (1 + x)dx

∴ tan–1 y = x + (x2/2) + c

at x = 0, y = 0, hence c = 0

∴ tan–1 y = x + (x2/2)

∴ y = tan[x + (x2/2)]