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

Solved using Taylor series

Another example using Taylor series method.

\begin{align*} y^{\prime }+2xy & =1+x+x^{2}\\ y^{\prime } & =1+x+x^{2}-2xy\\ & =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\\ F_{1} & =\left ( \frac {\partial F_{0}}{\partial x}\right ) +\left ( \frac {\partial F_{0}}{\partial y}\right ) F_{0}\\ & =1+2x-2y+\left ( -2x\right ) \left ( 1+x+x^{2}-2xy\right ) \\ & =4x^{2}y-2y-2x^{2}-2x^{3}+1\\ F_{2} & =\left ( \frac {\partial F_{1}}{\partial x}\right ) +\left ( \frac {\partial F_{1}}{\partial y}\right ) F_{0}\\ & =\left ( 8xy-4x-6x^{2}\right ) +\left ( 4x^{2}-2\right ) \left ( x-2xy\right ) \\ & =12xy-8x^{3}y-6x-6x^{2}+4x^{3}\\ F_{3} & =\left ( \frac {\partial F_{2}}{\partial x}\right ) +\left ( \frac {\partial F_{2}}{\partial y}\right ) F_{0}\\ & =12y-24x^{2}y-6-12x+12x^{2}+\left ( 12x-8x^{3}\right ) \left ( 1+x+x^{2}-2xy\right ) \\ & =12y-48x^{2}y+16x^{4}y+24x^{2}+4x^{3}-8x^{4}-8x^{5}-6 \end{align*}

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

\begin{align*} F_{0} & =1\\ F_{1} & =-2y_{0}+1\\ F_{2} & =0\\ F_{3} & =12y_{0}-6 \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}+1\right ) \frac {x^{2}}{2}+\left ( 12y_{0}-6\right ) \frac {x^{4}}{24}+\cdots \\ & =y_{0}\left ( 1-x^{2}+\frac {1}{2}x^{4}+\cdots \right ) +\left ( x+\frac {1}{2}x^{2}-\frac {1}{4}x^{4}+\cdots \right ) \end{align*}