Problem 2. (16 pts) solve the initial value problem dy dx. This is a linear equation; multiply by the integrating factor = e 3x to obtain. 3e 3xy = 2xe 3x; e 3x dy dx e 3x dy dx d(ye 3x) dx. = 2xe 3x. ye 3x = z 2xe 3xdx = . 9 e 3x + c, hence the general solution is y = ce3x . Since y(0) = 0, c = 2/9 and y = Problem 3. (16 pts) solve the initial value problem dy dx sin x y x cos y y(0) = 0. Since (sin x y)dx + (cos y x)dy = 0. Substituting this formula into the second equation gives. X + (y) = cos y x, hence. (y) = cos y, (y) = sin y, f = sin y cos x xy and the general solution is given implicitly by sin y cos x xy = c.
