2.2.1.1 Example 1 \(y^{\prime }=B+Cf\left ( ax+by+c\right ) \)

Solve

\begin{equation} y^{\prime }=B+Cf\left ( ax+by+c\right ) \tag {1}\end{equation}

This form of ode can be solved by letting \(u=ax+by+c\) which makes the ode separable.

\[ \frac {du}{dx}=a+by^{\prime }\]

Or

\[ y^{\prime }=\frac {u^{\prime }-a}{b}\]

The ode becomes

\begin{align*} \frac {u^{\prime }-a}{b} & =B+CF\left ( u\right ) \\ u^{\prime } & =bB+bCF\left ( u\right ) +a\\ \frac {du}{bB+bCF\left ( u\right ) +a} & =dx \end{align*}

Integrating gives

\begin{equation} \int \frac {du}{bB+bCF\left ( u\right ) +a}=x+c \tag {2A}\end{equation}

If initial conditions are given as \(y\left ( x_{0}\right ) =y_{0}\), then \(c\) is solved for using these values if it is possible to integrate 2A. If not possible to integrate (2A), then we can do the following. Rewrite (2A) as

\begin{align} \int _{0}^{ax_{0}+by_{0}+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a} & =x_{0}+c_{1}\tag {2B}\\ c_{1} & =\int _{0}^{ax_{0}+by_{0}+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a}-x_{0}\nonumber \end{align}

Substituting this into (2A) gives

\[ \int _{0}^{ax+by+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a}=x+\int _{0}^{ax_{0}+by_{0}+c}\frac {d\tau }{bB+bCF\left ( \tau \right ) +a}-x_{0}\]

Note that when IC are given, the integrals are changed to have lower limit start from zero.