Question

dy/dx=1+tan(y-x) put y-x=z​

Answer 1

y(x) = (3 ∫ x. 0 et2 dt + 1). 1/3 . (d) dy dx. = (1 + y2)√1 + sin x =⇒ dy. 1 + y2= √1 + sin x dx =⇒ tan−1 y = −2. √1. − sin x + C. y(0)=1 =⇒ C = π. 4. + 2. Therefore: y(x) ... Interval of definition. By looking at an initial value problem dy/dx = f(x, y) with y(x0) = y0, ... D(y(x)) = lim a→+∞(−1/a,1/a)=(−0,0) = {}; i.e. the empty set. 4.