2.20.3.4.1 Examples for section 3-1-3 (a) Removal of linear term, subcase (i)

2.20.3.4.1.1 Algorithm Given \(y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\) let \(y=u\left ( x\right ) e^{\phi }\) where \(\phi =\int f_{1}dx\) and the ode becomes

\[ u^{\prime }=F\left ( x\right ) +G\left ( x\right ) u^{2}\]

Where \(F\left ( x\right ) =f_{0}e^{-\phi },G\left ( x\right ) =f_{2}e^{\phi }\). There is one special subcase here to consider. If \(F\left ( x\right ) \) turns out to be proportional to \(G\left ( x\right ) \) then the ode is separable. So this check should be done after completing the above. See second example below of one such example.

2.20.3.4.1.2 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.

2.20.3.4.1.3 Example \(y^{\prime }=5xe^{3x}+3y+10xe^{-3x}y^{2}\) Comparing to \(y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\) shows that

\begin{align*} f_{0} & =5xe^{3x}\\ f_{1} & =3\\ f_{2} & =10xe^{-3x}\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

\[ u^{\prime }=F\left ( x\right ) +G\left ( x\right ) u^{2}\]

Where

\begin{align*} F\left ( x\right ) & =f_{0}e^{-\phi }=5xe^{3x}e^{-3x}=5x\\ G\left ( x\right ) & =f_{2}e^{\phi }=10xe^{-3x}e^{3x}=10x \end{align*}

Hence

\[ u^{\prime }=5x+10xu^{2}\]

Since \(F\left ( x\right ) \) is proportional to \(G\left ( x\right ) \), then this is separable ode which is easily solved. Once \(u\) is known, then \(y\) is found since \(y=u\left ( x\right ) e^{\phi }\).