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: $x = -\frac{8}{\pi}$

Explanation: Note that

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

To evaluate the integral in the boxed equation, let $u = \frac{\pi y}{4}$. Then, 

$du = \frac{\pi}{4} dy \\ \Rightarrow \boxed{dy = \frac{4}{\pi}du}$ 

So,

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

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

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

With the value of C, equation (1) becomes

$\boxed{x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) -\frac{4}{\pi} }$

Hence, if y = 3, 

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