Answer
The solution of this differential equation is $\tan{(2y)}-\sin{(2x)}-x=C_0$.
Work Step by Step
We have given $\dfrac{dy}{dx}=(\cos^2{x})(\cos^2{2y})$
Now solve the differential equation as follows:
$\dfrac{dy}{dx}=(\cos^2{x})(\cos^2{2y})$
$\dfrac{dy}{\cos^2{(2y)}}=(\cos^2{x})dx$
Since, $\dfrac{1}{\cos{x}}=\sec{x}$
We get, $\sec^2{(2y)}\,dy=(\cos^2{x})dx$
Substitute, $\cos^2{x}=\dfrac{\cos{2x}+1}{2}$
We get, $\sec^2{(2y)}\,dy=\left(\dfrac{\cos{2x}+1}{2}\right)dx$
Now integrate both sides as follows:
$\int\sec^2{(2y)}\,dy=\int\dfrac{\cos{2x}}{2}dx+\int\dfrac{1}{2}dx$
We get, $\dfrac{\tan{(2y)}}{2}=\dfrac{\sin{(2x)}}{2}+\dfrac{x}{2}+C$
Rearrange the equation
We get, $\dfrac{\tan{(2y)}}{2}-\dfrac{\sin{(2x)}}{2}-\dfrac{x}{2}=C$
Or, $\dfrac{\tan{(2y)}-\sin{(2x)}-x}{2}=C$
Or, $\tan{(2y)}-\sin{(2x)}-x=C_0$
Hence, the solution of this differential equation is $\tan{(2y)}-\sin{(2x)}-x=C_0$.
