Differential Equations

Question

For the following differential equation, given below, find particular solution satisfying the given condition:
$cos space open parentheses dy over dx close parentheses space space equals straight a space space space left parenthesis straight a space element of space straight R right parenthesis semicolon space space space straight y space equals space 1 space space when space straight x space equals space 0$

Solution

The given differential equation is

cos space open parentheses dy over dx close parentheses space equals space straight a space space space space or space space space dy over dx space equals space cos to the power of negative 1 end exponent straight a

therefore space space space space space space space space space dy space equals space cos to the power of negative 1 end exponent straight a. space dx space space space rightwards double arrow space space space space integral space dy space equals space cos to the power of negative 1 end exponent space straight a integral space dx

therefore space space space space space space space space straight y space equals space left parenthesis cos to the power of negative 1 end exponent straight a right parenthesis. space straight x plus space straight c

...(1)
Now y = 1 when x = 0

therefore space space space 1 space equals space left parenthesis cos to the power of negative 1 end exponent straight a right parenthesis space. space 0 space plus straight c space rightwards double arrow space space 1 space equals space straight c therefore space space space space from space left parenthesis 1 right parenthesis comma space space straight y space equals space straight x space cos to the power of negative 1 end exponent straight a plus 1