Answer to Question #130749 in Differential Equations for Nikhil

Question

Solve the following equation
sin(y')=0

Answer 1

sin (y'(x)) = 0$\implies$ The values for which sine function gives 0 value are n$\pi$ , where n is an integer. So, here we have y'(x)= n$\pi$
=> y(x) = $\int$ n$\pi$ dx + C , where C is an integration constant.=> y(x) = n$\pi$ x + C , which is the required answer to the given differential equation.