This help page gives a few examples of using the command ODESteps to solve ordinary differential equations by means of series expansions.

See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence.

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right)\:$

$\mathrm{ode1}≔x^2\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+x\mathrm{diff}\left(y\left(x\right)\,x\right)+5xy\left(x\right)=0$

$\mathrm{ode1}≔x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+5xy\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode1}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+5xy\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=-\frac{5y\left(x\right)}{x}-\frac{\frac{\ⅆ}{\ⅆx}y\left(x\right)}{x}\\ \text{•}& & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+\frac{\frac{\ⅆ}{\ⅆx}y\left(x\right)}{x}+\frac{5y\left(x\right)}{x}=0\\ \text{▫}& & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & \text{◦}& \text{Define functions}\\ & & \left[P_2\left(x\right)=\frac{1}{x}\,P_3\left(x\right)=\frac{5}{x}\right]\\ & \text{◦}& P_2\left(x\right)\text{is analytic at}x=0\\ & & \left[\right]=1\\ & \text{◦}& P_3\left(x\right)\text{is analytic at}x=0\\ & & \left[\right]=0\\ & \text{◦}& x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0=0\text{is a regular singular point}\\ & & x_0=0\\ \text{•}& & \text{Multiply by denominators}\\ & & \left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)x+5y\left(x\right)+\frac{\ⅆ}{\ⅆx}y\left(x\right)=0\\ \text{•}& & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^{k+r}\\ \text{▫}& & \text{Rewrite ODE with series expansions}\\ & \text{◦}& \text{Convert}\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)x\text{to series expansion}\\ & & \left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)x=\sum _{k=0}^{\mathrm{\infty }}{a}_{k+1}\left(k+r+1\right)\left(k+r\right)x^{k+r}\\ & \text{◦}& \text{Convert}5y\left(x\right)\text{to series expansion}\\ & & 5y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}5a_k{x}^{k+r}\\ & \text{◦}& \text{Convert}\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{to series expansion}\\ & & \frac{\ⅆ}{\ⅆx}y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k+1}\left(k+r+1\right)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & \mathrm{Sum}\left(\left(a_{k+1}\left(k+r+1\right)+a_{k+1}\left(k+r+1\right)\left(k+r\right)+5a_k\right)x^{k+r}\,0..\mathrm{\infty }\right)=0\\ \text{•}& & a_0\text{cannot be 0 by assumption giving the indicial equation}\\ & & r=0\\ \text{•}& & \text{Values of r that satisfy the indicial equation}\\ & & r=0\\ \text{•}& & \text{Each term must evaluate to 0 giving the recursion relation}\\ & & {\left(k+r+1\right)}^2{a}_{k+1}+5a_k\\ \text{•}& & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=-\frac{5a_k}{{\left(k+r+1\right)}^2}\\ \text{•}& & \text{Solution for}r=0\\ & & \left[y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k\,a_{k+1}=-\frac{5a_k}{{\left(k+1\right)}^2}\right]\end{array}$

$\mathrm{ode2}≔\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+x\mathrm{diff}\left(y\left(x\right)\,x\right)+y\left(x\right)=0$

$\mathrm{ode2}≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode2}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)+x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+y\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k\\ \text{▫}& & \text{Rewrite ODE with series expansions}\\ & \text{◦}& \text{Convert}x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\text{to series expansion}\\ & & x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k}k{x}^k\\ & \text{◦}& \text{Convert}\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\text{to series expansion}\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k+2}\left(k+2\right)\left(k+1\right)x^k\\ & \text{◦}& \text{Convert}y\left(x\right)\text{to series expansion}\\ & & y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k\\ & & \text{Rewrite ODE with series expansions}\\ & & \mathrm{Sum}\left(\left(a_k+a_{k}k+a_{k+2}\left(k+2\right)\left(k+1\right)\right)x^k\,0..\mathrm{\infty }\right)=0\\ \text{•}& & \text{Each term must evaluate to 0 giving the recursion relation}\\ & & \left(k+1\right)\left(\left(k+2\right)a_{k+2}+a_k\right)=0\\ \text{•}& & \text{Recursion relation that defines series solution to ODE}\\ & & \left[y\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k\,a_{k+2}=-\frac{a_k}{k+2}\right]\end{array}$

$\mathrm{ode3}≔x^2\mathrm{diff}\left(y\left(x\right)\,x\,x\right)+x^2\mathrm{diff}\left(y\left(x\right)\,x\right)+\left(x^3-6\right)y\left(x\right)=0$

$\mathrm{ode3}≔x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+x^2\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+\left(x^3-6\right)y\left(x\right)=0$

$\mathrm{ODESteps}\left(\mathrm{ode3}\right)$

$\begin{array}{lll}& & \text{Let's solve}\\ & & x^2\left(\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\right)+x^2\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+\left(x^3-6\right)y\left(x\right)=0\\ \text{•}& & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\\ \text{•}& & \text{Isolate 2nd derivative}\\ & & \frac{{\ⅆ}^2}{\ⅆx^{}}\end{array}$