ODE Steps for Series Solutions
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.
Let's solvex2⁢ⅆ2ⅆx2y⁡x+x⁢ⅆⅆxy⁡x+5⁢x⁢y⁡x=0•Highest derivative means the order of the ODE is2ⅆ2ⅆx2y⁡x•Isolate 2nd derivativeⅆ2ⅆx2y⁡x=−5⁢y⁡xx−ⅆⅆxy⁡xx•Group terms withy⁡xon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2y⁡x+ⅆⅆxy⁡xx+5⁢y⁡xx=0▫Check to see ifx0=0is a regular singular point◦Define functionsP2⁡x=1x,P3⁡x=5x◦P2⁡xis analytic atx=0=1◦P3⁡xis analytic atx=0=0◦x=0is a regular singular pointCheck to see ifx0=0is a regular singular pointx0=0•Multiply by denominatorsⅆ2ⅆx2y⁡x⁢x+5⁢y⁡x+ⅆⅆxy⁡x=0•Assume series solution fory⁡xy⁡x=∑k=0∞⁡ak⁢xk+r▫Rewrite ODE with series expansions◦Convertⅆ2ⅆx2y⁡x⁢xto series expansionⅆ2ⅆx2y⁡x⁢x=∑k=0∞⁡ak+1⁢k+r+1⁢k+r⁢xk+r◦Convert5⁢y⁡xto series expansion5⁢y⁡x=∑k=0∞⁡5⁢ak⁢xk+r◦Convertⅆⅆxy⁡xto series expansionⅆⅆxy⁡x=∑k=0∞⁡ak+1⁢k+r+1⁢xk+rRewrite ODE with series expansionsSum⁡ak+1⁢k+r+1+ak+1⁢k+r+1⁢k+r+5⁢ak⁢xk+r,0..∞=0•a0cannot be 0 by assumption giving the indicial equationr=0•Values of r that satisfy the indicial equationr=0•Each term must evaluate to 0 giving the recursion relationk+r+12⁢ak+1+5⁢ak•Recursion relation that defines series solution to ODEak+1=−5⁢akk+r+12•Solution forr=0y⁡x=∑k=0∞⁡ak⁢xk,ak+1=−5⁢akk+12
Let's solveⅆ2ⅆx2y⁡x+x⁢ⅆⅆxy⁡x+y⁡x=0•Highest derivative means the order of the ODE is2ⅆ2ⅆx2y⁡x•Assume series solution fory⁡xy⁡x=∑k=0∞⁡ak⁢xk▫Rewrite ODE with series expansions◦Convertx⁢ⅆⅆxy⁡xto series expansionx⁢ⅆⅆxy⁡x=∑k=0∞⁡ak⁢k⁢xk◦Convertⅆ2ⅆx2y⁡xto series expansionⅆ2ⅆx2y⁡x=∑k=0∞⁡ak+2⁢k+2⁢k+1⁢xk◦Converty⁡xto series expansiony⁡x=∑k=0∞⁡ak⁢xkRewrite ODE with series expansionsSum⁡ak+ak⁢k+ak+2⁢k+2⁢k+1⁢xk,0..∞=0•Each term must evaluate to 0 giving the recursion relationk+1⁢k+2⁢ak+2+ak=0•Recursion relation that defines series solution to ODEy⁡x=∑k=0∞⁡ak⁢xk,ak+2=−akk+2