1.2.2.3 Example 3 \(y^{\prime }+2xy^{2}=1+x+x^{2}\)

Solved using Taylor series

\begin{align*} y^{\prime }+2xy^{2} & =1+x+x^{2}\\ y^{\prime } & =1+x+x^{2}-2xy^{2}\\ & =f\left ( x,y\right ) \end{align*}

Let expansion point be \(x=0\). Let \(y\left ( 0\right ) =y_{0}\). Then

\[ y=y\left ( 0\right ) +\sum _{n=0}^{\infty }\frac {x^{n+1}}{\left ( n+1\right ) !}\left . F_{n}\left ( x,y\right ) \right \vert _{x=0,y_{0}}\]

Where \(F_{0}=f\left ( x,y\right ) \) and \(F_{n}=\frac {\partial F_{n-1}}{\partial x}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) F_{0}\). Hence

\begin{align*} F_{0} & =1+x+x^{2}-2xy^{2}\\ F_{1} & =\left ( 1+2x-2y^{2}\right ) +\left ( -4xy\right ) \left ( 1+x+x^{2}-2xy^{2}\right ) \\ & =-4x^{3}y+8x^{2}y^{3}-4x^{2}y-4xy+2x-2y^{2}+1\\ F_{2} & =\left ( \frac {\partial F_{1}}{\partial x}\right ) +\left ( \frac {\partial F_{1}}{\partial y}\right ) F_{0}\\ & =\left ( -12x^{2}y+16xy^{3}-8xy-4y+2\right ) +\left ( -4x^{3}+24x^{2}y^{2}-4x^{2}-4x-4y\right ) \left ( 1+x+x^{2}-2xy^{2}\right ) \\ & =-4x^{5}+32x^{4}y^{2}-8x^{4}-48x^{3}y^{4}+32x^{3}y^{2}-\allowbreak 12x^{3}+32x^{2}y^{2}-16x^{2}y-8x^{2}+24xy^{3}-12x\allowbreak y-4x-8y+2\\ F_{3} & =\left ( \frac {\partial F_{2}}{\partial x}\right ) +\left ( \frac {\partial F_{2}}{\partial y}\right ) F_{0}\end{align*}

And so on. Evaluating the above at \(x=0,y=y_{0}\) gives

\begin{align*} F_{0} & =1\\ F_{1} & =-2y_{0}^{2}+1\\ F_{2} & =-8y_{0}+2 \end{align*}

Hence

\begin{align*} y & =y\left ( 0\right ) +\sum _{n=0}^{\infty }\frac {x^{n+1}}{\left ( n+1\right ) !}\left . F_{n}\left ( x,y\right ) \right \vert _{x=0,y_{0}}\\ & =y_{0}+F_{0}x+F_{1}\frac {x^{2}}{2}+F_{2}\frac {x^{3}}{6}+F_{3}\frac {x^{4}}{24}+\cdots \\ & =y_{0}+x+\left ( -2y_{0}^{2}+1\right ) \frac {x^{2}}{2}+\left ( -8y_{0}+2\right ) \frac {x^{3}}{6}+\cdots \\ & =y_{0}\left ( 1-\frac {4}{3}x^{3}+\cdots \right ) +y_{0}^{2}\left ( -x^{2}+\cdots \right ) +\cdots +\left ( x+\frac {1}{2}x^{2}+\frac {1}{3}x^{3}+\cdots \right ) \end{align*}