Example 6 y′ = sin (x − y)

\begin{align*} y^{\prime } & =\sin \left ( x-y\right ) \\ y\left ( 1\right ) & =1 \end{align*}

Comparing the above to (1) which is \(y^{\prime }=B+Cf\left ( ax+by+c\right ) \) shows that

\begin{align*} B & =0\\ C & =1\\ a & =1\\ b & =-1\\ c & =0 \end{align*}

Using (2A) above

\[ \int \frac {du}{bB+bCF\left ( u\right ) +a}=x+c \]

Then the above becomes, where here \(F\left ( u\right ) =\sin \left ( u\right ) \)

\begin{align*} \int \frac {du}{-\sin \left ( u\right ) +1} & =x+c\\ \frac {2}{1-\tan \left ( \frac {u}{2}\right ) } & =x+c \end{align*}

But \(u=x-y\), therefore

\begin{equation} \frac {2}{1-\tan \left ( \frac {x-y}{2}\right ) }=x+c \tag {1}\end{equation}

Since initial conditions are given as \(y\left ( x_{0}\right ) =y_{0}\), where \(x_{0}=1,y_{0}=1\) then the above becomes

\begin{align*} \frac {2}{1-\tan \left ( \frac {0}{2}\right ) } & =1+c\\ c & =1 \end{align*}

Therefore solution (1) becomes

\[ \frac {2}{1-\tan \left ( \frac {x-y}{2}\right ) }=x+1 \]

Solving for \(y\) gives

\[ y(x)=x-2\arctan \left ( \frac {x-1}{1+x}\right ) \]