Given, (2xy dx + dy) e^(x^2) = 0, with the initial condition y(0) = 2. If exact, solve it or if not, then by using the integrating factors solve it and find the particular solution.

Answer 1

e^{x^2}≠ 0 for any x. So, 2xydx + dy = 0; dy/dx = - 2xy; dy/y = - 2xdx; lny = - x² + lnC; y = Ce^{-x^2}

y(0) = C = 2; So, y = 2e^{-x^2}