2.2.1.2 Example 2 \(y^{\prime }=\frac {1}{7}F\left ( 3x+5y\right ) \)

Solve

\begin{align*} y^{\prime } & =\frac {1}{7}F\left ( 3x+5y\right ) \\ y\left ( x_{0}\right ) & =y_{0}\end{align*}

Comparing the above to (1) shows that

\begin{align*} B & =0\\ C & =\frac {1}{7}\\ a & =3\\ b & =5\\ c & =0 \end{align*}

Plugging these into (2B) gives

\begin{align*} \int ^{ax+by+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a} & =x+c_{1}\\ \int ^{3x+5y}\frac {d\tau }{\frac {5}{7}F\left ( \tau \right ) +3} & =x+c_{1}\end{align*}

Applying IC gives

\[ \int _{0}^{3x_{0}+5y_{0}}\frac {d\tau }{\frac {5}{7}F\left ( \tau \right ) +3}=c_{1}\]

Hence the solution is

\[ \int _{0}^{3x+5y}\frac {d\tau }{\frac {5}{7}F\left ( \tau \right ) +3}=x+\int _{0}^{3x_{0}+5y_{0}}\frac {d\tau }{\frac {5}{7}F\left ( \tau \right ) +3}\]

If IC were given as \(y\left ( 0\right ) =0\) then we see that \(c_{1}=0\) because upper limit becomes zero and the above solution becomes

\[ \int _{0}^{3x+5y}\frac {d\tau }{\frac {5}{7}F\left ( \tau \right ) +3}=x \]