2.20.3.4.3 Examples for section 3-1-3 (a) Removal of linear term, subcase (iii)

2.20.3.4.3.1 Example \(y^{\prime }=x+3y+7y^{2}\) Comparing to \(y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\) shows that

\begin{align*} f_{0} & =x\\ f_{1} & =3\\ f_{2} & =7 \end{align*}

Let

\begin{align*} y & =u\left ( z\right ) e^{\phi }\\ \phi & =\int f_{1}dx\\ z & =-\int f_{2}e^{\phi }dx \end{align*}

Hence \(\phi =\int 3dx=3x\). Applying this transformation results in new ode in \(u\left ( z\right ) \) as

\[ u^{\prime }\left ( z\right ) =F\left ( z\right ) -u^{2}\left ( z\right ) \]

Where

\begin{align*} F\left ( z\right ) & =-\frac {f_{0}\left ( z\right ) e^{-2\phi }}{f_{2}\left ( z\right ) }\\ & =\frac {-x}{7}e^{-6x}\end{align*}

But \(z=-\int f_{2}e^{\phi }dx=-\int 7e^{3x}dx=-\frac {7}{3}e^{3x}\). Hence \(-\frac {3z}{7}=e^{3x}\) or \(\ln \left ( -\frac {3z}{7}\right ) =3x\) or

\[ x=\frac {1}{3}\ln \left ( -\frac {3z}{7}\right ) \]

Therefore

\begin{align*} F\left ( z\right ) & =\frac {-1}{7}\frac {1}{3}\ln \left ( -\frac {3z}{7}\right ) \left ( e^{3x}\right ) ^{-2}\\ & =\frac {-1}{21}\ln \left ( -\frac {3z}{7}\right ) \left ( -\frac {3z}{7}\right ) ^{-2}\\ & =\frac {-1}{21}\ln \left ( -\frac {3z}{7}\right ) \left ( \frac {7^{2}}{9z^{2}}\right ) \\ & =\frac {-7}{3}\ln \left ( -\frac {3z}{7}\right ) \left ( \frac {1}{9z^{2}}\right ) \\ & =\frac {-7}{27z^{2}}\ln \left ( -\frac {3z}{7}\right ) \end{align*}

The new ode in \(u\) becomes

\begin{equation} u^{\prime }=\frac {-7}{27z^{2}}\ln \left ( -\frac {3z}{7}\right ) -u^{2} \tag {1}\end{equation}

We see that the linear term is removed. In the above \(f_{0}=\frac {-7}{27z^{2}}\ln \left ( -\frac {3z}{7}\right ) ,f_{1}=0,f_{2}=-1\).

Now we will try to solve (1). It is not one of the special cases we have solved before. We can either try to find a particular solution, or if we can’t, then convert it to second order ode and solve it that way.