Q.Solve the differential equation $\cos(x + y)\,dy = dx$. Hence find the particular solution for $x = 0$ and $y = 0$.

Substitute $u=x+y$ to reduce to a separable equation in $u$ and $x$, using $1+\cos u=2\cos^2(u/2)$.

$\cos(x+y)\,dy=dx \Rightarrow \frac{dy}{dx}=\frac{1}{\cos(x+y)}=\sec(x+y)$

Let $u=x+y$, so $\dfrac{du}{dx}=1+\dfrac{dy}{dx}$, i.e. $\dfrac{dy}{dx}=\dfrac{du}{dx}-1$.

$\frac{du}{dx}-1=\sec u \Rightarrow \frac{du}{dx}=1+\sec u$

Separate variables:

$\frac{du}{1+\sec u}=dx \Rightarrow \frac{\cos u}{1+\cos u}\,du=dx$

Using $1+\cos u=2\cos^2(u/2)$ and $\cos u=2\cos^2(u/2)-1$:

$\frac{\cos u}{1+\cos u}=\frac{2\cos^2(u/2)-1}{2\cos^2(u/2)}=1-\frac12\sec^2\left(\frac u2\right)$

Integrate both sides:

$\int\left[1-\frac12\sec^2\left(\frac u2\right)\right]du=\int dx$

$u-\frac12\cdot2\tan\left(\frac u2\right)=x+C$

$u-\tan\left(\frac u2\right)=x+C$

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