4.2.1.3 Example 3 \(y^{\prime }=\sin \left ( 3x+5y\right ) \)

\begin{align*} y^{\prime } & =\sin \left ( 3x+5y\right ) \\ y\left ( \frac {1}{3}\right ) & =-\frac {1}{5}\end{align*}

Comparing the above to (1) shows that

\begin{align*} B & =0\\ C & =1\\ a & =3\\ b & =5\\ 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}{5\sin \left ( u\right ) +3} & =x+c\\ \frac {1}{4}\ln \left ( 3\tan \frac {u}{2}+1\right ) -\frac {1}{4}\ln \left ( 3+\tan \frac {u}{2}\right ) & =x+c \end{align*}

But \(u=3x+5y\), therefore

\begin{equation} \frac {1}{4}\ln \left ( 3\tan \frac {3x+5y}{2}+1\right ) -\frac {1}{4}\ln \left ( 3+\tan \frac {3x+5y}{2}\right ) =x+c \tag {1}\end{equation}

Since initial conditions are given as \(y\left ( x_{0}\right ) =y_{0}\), where \(x_{0}=\frac {1}{3},y_{0}=-\frac {1}{5}\) then the above becomes

\begin{align*} -\frac {1}{4}\ln \left ( 3\tan \frac {0}{2}+1\right ) -\frac {1}{4}\ln \left ( 3+\tan \frac {0}{2}\right ) & =\frac {1}{3}+c\\ -\frac {1}{4}\ln \left ( 1\right ) -\frac {1}{4}\ln \left ( 3\right ) & =\frac {1}{3}+c\\ -\frac {1}{4}\ln \left ( 3\right ) & =\frac {1}{3}+c\\ c & =-\frac {1}{4}\ln \left ( 3\right ) -\frac {1}{3}\end{align*}

Hence solution (1) becomes

\[ \frac {1}{4}\ln \left ( 3\tan \frac {3x+5y}{2}+1\right ) -\frac {1}{4}\ln \left ( 3+\tan \frac {3x+5y}{2}\right ) =x-\frac {1}{4}\ln \left ( 3\right ) -\frac {1}{3}\]

Solving the above for \(y\) gives

\[ y(x)=-\frac {3}{5}x-\frac {2}{5}\arctan \left ( \frac {3\left ( -1+e^{-\frac {4}{3}+4x}\right ) }{e^{-\frac {4}{3}+4x}-9}\right ) \]