Question

If dy dx equals cosine squared of the quantity pi times y over 4 and y = 1 when x = 0, then find the value of x when y = 3.

Answer 1

Answer:
`x = -(8)/(\pi)`

Explanation: Note that


`(dy)/(dx) = \cos^2 \left ( (\pi y)/(4) \right ) \\ \\ (dy)/(\cos^2 \left ( (\pi y)/(4) \right )) = dx \\ \\ \int{(dy)/(\cos^2 \left ( (\pi y)/(4) \right ))} = \int dx \\ \\ \boxed{x = \int{\sec^2 \left ( (\pi y)/(4) \right )dy}}`

To evaluate the integral in the boxed equation, let
`u = (\pi y)/(4)`. Then,


`du = (\pi)/(4) dy \\ \Rightarrow \boxed{dy = (4)/(\pi)du}`

So,


`x = \int{\sec^2 \left ( (\pi y)/(4) \right )dy} \\ \\ = \int{\sec^2 u \left ( (4)/(\pi)du \right )} \\ \\ = (4)/(\pi)\int{\sec^2 u du} \\ \\ = (4)/(\pi) \tan u + C \\ \\ \boxed{x = (4)/(\pi) \tan \left ( (\pi y)/(4) \right ) + C}\text{  (1)}`

Since y = 1 when x =0, equation (1) becomes


`0 = (4)/(\pi) \tan \left ( (\pi (1))/(4) \right ) + C \\ \\ 0 = (4)/(\pi) (1) + C \\ \\ (4)/(\pi) + C = 0 \\ \\ \boxed{C = -(4)/(\pi)}`

With the value of C, equation (1) becomes


`\boxed{x = (4)/(\pi) \tan \left ( (\pi y)/(4) \right ) -(4)/(\pi) }`

Hence, if y = 3,


`x = (4)/(\pi) \tan \left ( (\pi y)/(4) \right ) -(4)/(\pi) \\ \\ x = (4)/(\pi) \tan \left ( (\pi (3))/(4) \right ) -(4)/(\pi) \\ \\ x = (4)/(\pi) (-1) -(4)/(\pi) \\ \\ \boxed{x = -(8)/(\pi)}`