4.2.1.2 Example 2 $y^{\prime }=\frac{1}{7}F(3x+5y)$

Solve

$$\begin{array}{rl}{y}^{\prime }& =\frac{1}{7}F(3x+5y)\\ y\left(x_0\right)& =y_0\end{array}$$

Comparing the above to (1) shows that

$$\begin{array}{rl}B& =0\\ C& =\frac{1}{7}\\ a& =3\\ b& =5\\ c& =0\end{array}$$

Plugging these into (2B) gives

$$\begin{array}{rl}{\int }^{ax+by+c}\frac{d\tau }{bB+bCF\left(\tau \right)+a}& =x+c_1\\ {\int }^{3x+5y}\frac{d\tau }{\frac{5}{7}F\left(\tau \right)+3}& =x+c_1\end{array}$$

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

$${\int }_0^{3x+5y}\frac{d\tau }{\frac{5}{7}F\left(\tau \right)+3}=x$$