Answer :
Answer: [tex]x = -\frac{8}{\pi}[/tex]
Explanation: Note that
[tex]\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}}[/tex]
To evaluate the integral in the boxed equation, let [tex]u = \frac{\pi y}{4}[/tex]. Then,
[tex]du = \frac{\pi}{4} dy \\ \Rightarrow \boxed{dy = \frac{4}{\pi}du}[/tex]
So,
[tex]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)}[/tex]
Since y = 1 when x =0, equation (1) becomes
[tex]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}}[/tex]
With the value of C, equation (1) becomes
[tex]\boxed{x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) -\frac{4}{\pi} }[/tex]
Hence, if y = 3,
[tex]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}} [/tex]
Explanation: Note that
[tex]\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}}[/tex]
To evaluate the integral in the boxed equation, let [tex]u = \frac{\pi y}{4}[/tex]. Then,
[tex]du = \frac{\pi}{4} dy \\ \Rightarrow \boxed{dy = \frac{4}{\pi}du}[/tex]
So,
[tex]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)}[/tex]
Since y = 1 when x =0, equation (1) becomes
[tex]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}}[/tex]
With the value of C, equation (1) becomes
[tex]\boxed{x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) -\frac{4}{\pi} }[/tex]
Hence, if y = 3,
[tex]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}} [/tex]