3.3.25.4 Ordinary point using Taylor series method

ode internal name "first_order_ode_taylor_series_method_ordinary_point"             

Alternative method to solving the above example is given here which is to use the Taylor series method.  This is derived as follows.

Let

$$y^{\prime }=f(x,y)$$

Where $f(x,y)$ is analytic at expansion point $x_0$. We can always shift to $x_0=0$ if $x_0$ is not zero. So from now we assume $x_0=0$. Assume also that $y\left(x_0\right)=y_0$. Using Taylor series

$$\begin{array}{rl}y\left(x\right)& =y\left(x_0\right)+(x-x_0)y^{\prime }\left(x_0\right)+\frac{{(x-x_0)}^2}{2}{y}^{\prime \prime }\left(x_0\right)+\frac{{(x-x_0)}^3}{3!}{y}^{\prime \prime }\left(x_0\right)+\cdots \\ & =y_0+xf+\frac{x^2}{2}{\frac{df}{dx}|}_{x_0,y_0}+\frac{x^3}{3!}{\frac{d^{2}f}{d{x}^2}|}_{x_0,y_0}+\cdots \\ & =y_0+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{(n+1)!}{\frac{d^{n}f}{d{x}^n}|}_{x_0,y_0}\end{array}$$

But

$$\begin{array}{rl}\text{(1)}& \frac{df}{dx}& =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}f\frac{d^{2}f}{d{x}^2}& =\frac{d}{dx}\left(\frac{df}{dx}\right)\\ \text{(2)}& & =\frac{\partial }{\partial x}\left(\frac{df}{dx}\right)+\frac{\partial }{\partial y}\left(\frac{df}{dx}\right)f\frac{d^{3}f}{d{x}^3}& =\frac{d}{dx}\left(\frac{d^{2}f}{d{x}^2}\right)\\ \text{(3)}& & =\frac{\partial }{\partial x}\left(\frac{d^{2}f}{d{x}^2}\right)+\left(\frac{\partial }{\partial y}\frac{d^{2}f}{d{x}^2}\right)f& ⋮\end{array}$$

And so on. Hence if we name $F_0=f(x,y)$ then the above can be written as

$$\begin{array}{rl}\text{(4)}& F_0& =f(x,y)F_n& =\frac{d}{dx}\left(F_{n-1}\right)\\ \text{(5)}& & =\frac{\partial }{\partial x}{F}_{n-1}+\left(\frac{\partial F_{n-1}}{\partial y}\right)F_0\end{array}$$

For example, for $n=1$ we see that

$$\begin{array}{rl}{F}_1& =\frac{d}{dx}\left(F_0\right)\\ & =\frac{\partial }{\partial x}{F}_0+\left(\frac{\partial F_0}{\partial y}\right)F_0\\ & =\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}f\end{array}$$

Which is (1). And when $n=2$

$$\begin{array}{rl}{F}_2& =\frac{d}{dx}\left(F_1\right)\\ & =\frac{\partial }{\partial x}{F}_1+\left(\frac{\partial F_1}{\partial y}\right)F_0\\ & =\frac{\partial }{\partial x}(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}f)+\frac{\partial }{\partial y}(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}f)f\\ & =\frac{\partial }{\partial x}\left(\frac{df}{dx}\right)+\frac{\partial }{\partial y}\left(\frac{df}{dx}\right)f\end{array}$$

Which is (2) and so on. Therefore (4,5) can be used from now on along with

$$\begin{array}{}\text{(6)}& y\left(x\right)=y_0+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{(n+1)!}{F_n|}_{x_0,y_0}\end{array}$$

See below for examples.

3.3.25.4.1 Example 1

$$y^{\prime }+2xy=x$$

Solved using power series

Expansion is around $x=0$. The (homogeneous) ode has the form $y^{\prime }+p\left(x\right)y=0$. We see that $p\left(x\right)$ is defined as is at $x=0$. Hence this is an ordinary point, also the RHS has series expansion at $x=0$. It is very important to check that the RHS has series expansion at $x=0$. Otherwise this method will fail and we must use Frobenius even if $x=0$ is ordinary point for the LHS of the ode. For example for the ode $y^{\prime }+2xy=\frac{1}{x}$ or $y^{\prime }+2xy=\sqrt{x}$ standard power series will fail. See examples below.

Using standard power series, let

$$\begin{array}{rl}y& =\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n\\ y^{\prime }& =\sum _{n=0}^{\mathrm{\infty }}n{a}_n{x}^{n-1}=\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}\end{array}$$

The ode now becomes

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}+2x\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n& =x\\ \sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}+\sum _{n=0}^{\mathrm{\infty }}2a_n{x}^{n+1}& =x\end{array}$$

Reindex so that all powers on $x$ are $n$ gives

$$\sum _{n=0}^{\mathrm{\infty }}(n+1)a_{n+1}{x}^n+\sum _{n=1}^{\mathrm{\infty }}2a_{n-1}{x}^n=x$$

For $n=0$, the RHS is zero, since there is no matching term with $x^0$, therefore the above gives

$$a_1=0$$

For $n=1$, the RHS is $x^1$ which gives

$$\begin{array}{rl}(n+1)a_{n+1}+2a_{n-1}& =1\\ 2a_2+2a_0& =1\\ a_2& =\frac{1-2a_0}{2}\end{array}$$

For $n\ge 2$ the RHS is zero and we have recurrence relation. Therefore we have

$$(n+1)a_{n+1}+2a_{n-1}=0$$

For $n=2$

$$\begin{array}{rl}3a_3+2a_1& =0\\ a_3& =-\frac{2a_1}{3}=0\end{array}$$

For $n=3$

$$\begin{array}{rl}4a_4+2a_2& =0\\ a_4& =-\frac{1}{2}{a}_2=-\frac{1}{2}\left(\frac{1-2a_0}{2}\right)=\frac{2a_0-1}{4}\end{array}$$

And so on. The solution is

$$\begin{array}{rl}y& =\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n\\ & =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots \\ & =a_0+\left(\frac{1-2a_0}{2}\right)x^2+\left(\frac{2a_0-1}{4}\right)x^4+\cdots \\ & =a_0(1-x^2+\frac{1}{2}{x}^4+\cdots )+(\frac{1}{2}{x}^2-\frac{1}{4}{x}^4+\cdots )\end{array}$$

Which can be written as

$$y=y\left(0\right)(1-x^2+\frac{1}{2}{x}^4+\cdots )+(\frac{1}{2}{x}^2-\frac{1}{4}{x}^4+\cdots )$$

Solved using Taylor series

$$\begin{array}{rl}{y}^{\prime }+2xy& =x\\ y^{\prime }& =x-2xy\\ & =f(x,y)\end{array}$$

For this method to work, $f(x,y)$ must be analytic at $x=x_0$, the expansion point.  Let expansion point be $x=0$. Let $y\left(0\right)=y_0$. Then

$$y=y\left(0\right)+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{(n+1)!}{F_n(x,y)|}_{x=0,y_0}$$

Where $F_0=f(x,y)$ and $F_n=\frac{\partial F_{n-1}}{\partial x}+\left(\frac{\partial F_{n-1}}{\partial y}\right)F_0$. Hence

$$\begin{array}{rl}{F}_0& =(x-2xy)\\ F_1& =\frac{d}{dx}{F}_0\\ & =\left(\frac{\partial F_0}{\partial x}\right)+\left(\frac{\partial F_0}{\partial y}\right)F_0\\ & =\left(\frac{\partial (x-2xy)}{\partial x}\right)+\left(\frac{\partial (x-2xy)}{\partial y}\right)(x-2xy)\\ & =(1-2y)-2x(x-2xy)\\ & =4x^{2}y-2y-2x^2+1\\ F_2& =\frac{d^2}{d{x}^2}{F}_1\\ & =\left(\frac{\partial F_1}{\partial x}\right)+\left(\frac{\partial F_1}{\partial y}\right)F_0\\ & =\left(\frac{\partial }{\partial x}(4x^{2}y-2y-2x^2+1)\right)+(\frac{\partial }{\partial y}4x^{2}y-2y-2x^2+1)(x-2xy)\\ & =(8xy-4x)+(4x^2-2)(x-2xy)\\ & =12xy-8x^{3}y-6x+4x^3\\ F_3& =\frac{d^3}{d{x}^3}{F}_2\\ & =\left(\frac{\partial F_2}{\partial x}\right)+\left(\frac{\partial F_2}{\partial y}\right)F_0\\ & =\left(\frac{\partial }{\partial x}(12xy-8x^{3}y-6x+4x^3)\right)+\left(\frac{\partial }{\partial y}(12xy-8x^{3}y-6x+4x^3)\right)(x-2xy)\\ & =12y-24x^{2}y-6+12x^2+(12x-8x^3)(x-2xy)\\ & =12y-48x^{2}y+16x^{4}y+24x^2-8x^4-6\end{array}$$

And so on. Evaluating the above at $x=0,y=y_0$ gives

$$\begin{array}{rl}{F}_0& =0\\ F_1& =-2y_0+1\\ F_2& =0\\ F_3& =12y_0-6\end{array}$$

Hence

$$\begin{array}{rl}y& =y\left(0\right)+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{(n+1)!}{F_n(x,y)|}_{x=0,y_0}\\ & =y_0+x{F}_0+\frac{x^2}{2}{F}_1+\frac{x^3}{6}{F}_2+\frac{x^4}{24}{F}_3+\cdots \\ & =y_0+0+\frac{x^2}{2}(-2y_0+1)+0+\frac{x^4}{24}(12y_0-6)+\cdots \\ & =y_0-2y_0\frac{x^2}{2}+\frac{x^2}{2}+\frac{1}{2}{y}_0{x}^4-\frac{x^4}{4}+\\ & =y_0(1-x^2+\frac{1}{2}{x}^4)+\frac{x^2}{2}-\frac{x^4}{4}+\cdots \end{array}$$

3.3.25.4.2 Example 2 Solved using Taylor series

Another example using Taylor series method.

$$\begin{array}{rl}{y}^{\prime }+2xy& =1+x+x^2\\ y^{\prime }& =1+x+x^2-2xy\\ & =f(x,y)\end{array}$$

Let expansion point be $x=0$. Let $y\left(0\right)=y_0$. Then

$$y=y\left(0\right)+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{(n+1)!}{F_n(x,y)|}_{x=0,y_0}$$

Where $F_0=f(x,y)$ and $F_n=\frac{\partial F_{n-1}}{\partial x}+\left(\frac{\partial F_{n-1}}{\partial y}\right)F_0$. Hence

$$\begin{array}{rl}{F}_0& =1+x+x^2-2xy\\ F_1& =\left(\frac{\partial F_0}{\partial x}\right)+\left(\frac{\partial F_0}{\partial y}\right)F_0\\ & =1+2x-2y+(-2x)(1+x+x^2-2xy)\\ & =4x^{2}y-2y-2x^2-2x^3+1\\ F_2& =\left(\frac{\partial F_1}{\partial x}\right)+\left(\frac{\partial F_1}{\partial y}\right)F_0\\ & =(8xy-4x-6x^2)+(4x^2-2)(x-2xy)\\ & =12xy-8x^{3}y-6x-6x^2+4x^3\\ F_3& =\left(\frac{\partial F_2}{\partial x}\right)+\left(\frac{\partial F_2}{\partial y}\right)F_0\\ & =12y-24x^{2}y-6-12x+12x^2+(12x-8x^3)(1+x+x^2-2xy)\\ & =12y-48x^{2}y+16x^{4}y+24x^2+4x^3-8x^4-8x^5-6\end{array}$$

And so on. Evaluating the above at $x=0,y=y_0$ gives

$$\begin{array}{rl}{F}_0& =1\\ F_1& =-2y_0+1\\ F_2& =0\\ F_3& =12y_0-6\end{array}$$

Hence

$$\begin{array}{rl}y& =y\left(0\right)+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{(n+1)!}{F_n(x,y)|}_{x=0,y_0}\\ & =y_0+F_{0}x+F_1\frac{x^2}{2}+F_2\frac{x^3}{6}+F_3\frac{x^4}{24}+\cdots \\ & =y_0+x+(-2y_0+1)\frac{x^2}{2}+(12y_0-6)\frac{x^4}{24}+\cdots \\ & =y_0(1-x^2+\frac{1}{2}{x}^4+\cdots )+(x+\frac{1}{2}{x}^2-\frac{1}{4}{x}^4+\cdots )\end{array}$$

3.3.25.4.3 Example 3 Solved using Taylor series

$$\begin{array}{rl}{y}^{\prime }+2x{y}^2& =1+x+x^2\\ y^{\prime }& =1+x+x^2-2x{y}^2\\ & =f(x,y)\end{array}$$

Let expansion point be $x=0$. Let $y\left(0\right)=y_0$. Then

$$y=y\left(0\right)+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{(n+1)!}{F_n(x,y)|}_{x=0,y_0}$$

Where $F_0=f(x,y)$ and $F_n=\frac{\partial F_{n-1}}{\partial x}+\left(\frac{\partial F_{n-1}}{\partial y}\right)F_0$. Hence

$$\begin{array}{rl}{F}_0& =1+x+x^2-2x{y}^2\\ F_1& =(1+2x-2y^2)+(-4xy)(1+x+x^2-2x{y}^2)\\ & =-4x^{3}y+8x^2{y}^3-4x^{2}y-4xy+2x-2y^2+1\\ F_2& =\left(\frac{\partial F_1}{\partial x}\right)+\left(\frac{\partial F_1}{\partial y}\right)F_0\\ & =(-12x^{2}y+16x{y}^3-8xy-4y+2)+(-4x^3+24x^2{y}^2-4x^2-4x-4y)(1+x+x^2-2x{y}^2)\\ & =-4x^5+32x^4{y}^2-8x^4-48x^3{y}^4+32x^3{y}^2-12x^3+32x^2{y}^2-16x^{2}y-8x^2+24x{y}^3-12xy-4x-8y+2\\ F_3& =\left(\frac{\partial F_2}{\partial x}\right)+\left(\frac{\partial F_2}{\partial y}\right)F_0\end{array}$$

And so on. Evaluating the above at $x=0,y=y_0$ gives

$$\begin{array}{rl}{F}_0& =1\\ F_1& =-2y_0^2+1\\ F_2& =-8y_0+2\end{array}$$

Hence

$$\begin{array}{rl}y& =y\left(0\right)+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{(n+1)!}{F_n(x,y)|}_{x=0,y_0}\\ & =y_0+F_{0}x+F_1\frac{x^2}{2}+F_2\frac{x^3}{6}+F_3\frac{x^4}{24}+\cdots \\ & =y_0+x+(-2y_0^2+1)\frac{x^2}{2}+(-8y_0+2)\frac{x^3}{6}+\cdots \\ & =y_0(1-\frac{4}{3}{x}^3+\cdots )+y_0^2(-x^2+\cdots )+\cdots +(x+\frac{1}{2}{x}^2+\frac{1}{3}{x}^3+\cdots )\end{array}$$

3.3.25.4.4 Example 4 Solved using power series

$$y^{\prime }+y=\sin x$$

Expansion is around $x=0$. The (homogenous) ode has the form $y^{\prime }+p\left(x\right)y=0$. We see that $p\left(x\right)$ is defined as is at $x=0$. Hence this is ordinary point, also the RHS has series expansion at $x=0$.

Let $y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n,y^{\prime }=\sum _{n=0}^{\mathrm{\infty }}n{a}_n{x}^{n-1}=\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}$. The ode becomes

$$\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=\sin x$$

Indexing so all powers of $x$ start at $n$ gives

$$\sum _{n=0}^{\mathrm{\infty }}(n+1)a_{n+1}{x}^n+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=\sin x$$

Expanding $\sin x$ in series gives

$$\sum _{n=0}^{\mathrm{\infty }}(n+1)a_{n+1}{x}^n+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$

For $n=0$, there is no term on RHS with $x^0$, hence we obtain

$$\begin{array}{rl}{a}_1+a_0& =0\\ a_1& =-a_0\end{array}$$

For $n=1$ there is one term $x^1$ on RHS, hence

$$\begin{array}{rl}2a_2+a_1& =1\\ a_2& =\frac{1-a_1}{2}=\frac{1+a_0}{2}\end{array}$$

For $n=2$ there is no term on RHS with $x^2$ hence

$$\begin{array}{rl}3a_3+a_2& =0\\ a_3& =-\frac{a_2}{3}=-\frac{\frac{1+a_0}{2}}{3}=-\frac{1}{6}{a}_0-\frac{1}{6}\end{array}$$

For $n=3$ there is term $-\frac{1}{6}{x}^3$ on RHS, hence

$$\begin{array}{rl}4a_4+a_3& =-\frac{1}{6}\\ a_4& =\frac{-\frac{1}{6}-a_3}{4}=\frac{-\frac{1}{6}-(-\frac{1}{6}{a}_0-\frac{1}{6})}{4}=\frac{1}{24}{a}_0\end{array}$$

And so on. The solution is

$$\begin{array}{rl}y& =\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n\\ & =a_0+a_{1}x+a_2{x}^2+\cdots \\ & =a_0-a_{0}x+\left(\frac{1+a_0}{2}\right)x^2+(-\frac{1}{6}{a}_0-\frac{1}{6})x^3+\left(\frac{1}{24}{a}_0\right)x^4+\cdots \\ & =a_0(1-x+\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\frac{1}{24}{x}^4-\cdots )+(\frac{1}{2}{x}^2-\frac{1}{6}{x}^3+\cdots )\end{array}$$