1.3.4.1.2 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 ( x\right ) e^{\phi }\\ \phi & =\int f_{1}dx \end{align*}

Therefore \(\phi =\int 3dx=3x\) and we have

\[ y=e^{3x}u\left ( x\right ) \]

Therefore

\[ y^{\prime }=3e^{3x}u+e^{3x}u^{\prime }\]

The input ode becomes

\begin{align} y^{\prime } & =x+3y+7y^{2}\nonumber \\ 3e^{3x}u+e^{3x}u^{\prime } & =x+3\left ( e^{3x}u\right ) +7\left ( e^{3x}u\right ) ^{2}\nonumber \\ e^{3x}u^{\prime } & =x+7e^{6x}u^{2}\nonumber \\ u^{\prime } & =xe^{-3x}+7e^{3x}u^{2} \tag {1}\end{align}

We see that the transformed \(u\) ode will always have the general form

\begin{align*} u^{\prime } & =F\left ( x\right ) +G\left ( x\right ) u^{2}\\ & =f_{0}e^{-\phi }+f_{2}e^{\phi }u^{2}\end{align*}

Where

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

Hence the linear term \(f_{1}\) is removed. 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.