Question

Solve:
$\cos (x+y)dy=dx$ given $y\left(0\right)=\left(0\right)$

Answer 1

$\cos (x+y)dy=dx$

Let $x+y=z\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1$

$\Rightarrow \cos \left(z\right)\frac{dy}{dx}=1$

$\Rightarrow \cos \left(z\right)(\frac{dz}{dx}-1)=1$

$\Rightarrow \frac{dz}{dx}=\sec z+1$

$\Rightarrow \int \frac{dz}{\sec z+1}=\int dx+c$

$\Rightarrow x+c=\int \frac{\cos xdz}{\cos z+1}$

$\Rightarrow x+c=\int \frac{(\cos x+1-1)dz}{\cos z+1}$

$\Rightarrow x+c=\int \frac{(\cos x+1)dz}{\cos z+1}-\int \frac{dz}{\cos z+1}$

$\Rightarrow x+c=z-\int \frac{dz}{2{\cos }^2\frac{z}{2}}$

$\Rightarrow x+c=z-\frac{1}{2}\int {\sec }^2\frac{z}{2}dz$

$\Rightarrow x+c=z-\frac{1}{2}\times 2\tan \frac{z}{2}$

$\Rightarrow x+c=x+y-\tan \left(\frac{x+y}{2}\right)$

$\Rightarrow y=\tan \left(\frac{x+y}{2}\right)+c$

When $x=0,y=0$

$\therefore c=0$

Hence, $y=\tan \left(\frac{x+y}{2}\right)$ is the solution of the given differential equation.