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ODE Steps for Special Function Solutions

 

Overview

Examples

Overview

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This help page gives a few examples of using the command ODESteps to solve ordinary differential equations in terms of special functions.

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

Examples

withStudent:-ODEs:

ode1x2diffyx,x,x+4xdiffyx,x+25x29yx=0

ode1x2ⅆ2ⅆx2yx+4xⅆⅆxyx+25x29yx=0

(1)

ODEStepsode1

Let's solvex2ⅆ2ⅆx2yx+4xⅆⅆxyx+25x29yx=0Highest derivative means the order of the ODE is2ⅆ2ⅆx2yxIsolate 2nd derivativeⅆ2ⅆx2yx=25x29yxx24ⅆⅆxyxxGroup terms withyxon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2yx+4ⅆⅆxyxx+25x29yxx2=0Simplify ODEx2ⅆ2ⅆx2yx+25yxx2+4xⅆⅆxyx9yx=0Make a change of variablest=5xComputeⅆⅆxyxⅆⅆxyx=5ⅆⅆtytCompute second derivativeⅆ2ⅆx2yx=25ⅆ2ⅆt2ytApply change of variables to the ODEt2ⅆ2ⅆt2yt+ytt2+4tⅆⅆtyt9yt=0Make a change of variablesyt=utt32Computeⅆⅆtytⅆⅆtyt=3ut2t52+ⅆⅆtutt32Computeⅆ2ⅆt2ytⅆ2ⅆt2yt=15ut4t723ⅆⅆtutt52+ⅆ2ⅆt2utt32Apply change of variables to the ODEutt2+ⅆ2ⅆt2utt2+ⅆⅆtutt45ut4=0ODE is now of the Bessel formSolution to Bessel ODEut=_C1BesselJ352,t+_C2BesselY352,tMake the change fromyxback toytyt=_C1BesselJ352,t+_C2BesselY352,tt32Make the change fromtback toxyx=_C1BesselJ352,5x+_C2BesselY352,5x525x32

(2)

ode2x2+1diffyx,x,xxdiffyx,x+yx=0

ode2x2+1ⅆ2ⅆx2yxxⅆⅆxyx+yx=0

(3)

ODEStepsode2

Let's solvex2+1ⅆ2ⅆx2yxxⅆⅆxyx+yx=0Highest derivative means the order of the ODE is2ⅆ2ⅆx2yxIsolate 2nd derivativeⅆ2ⅆx2yx=yxx21xⅆⅆxyxx21Group terms withyxon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2yxyxx21+xⅆⅆxyxx21=0Multiply by denominators of ODEx2+1ⅆ2ⅆx2yxxⅆⅆxyx+yx=0Make a change of variablesθ=arccosxCalculateⅆⅆxyxwith change of variablesⅆⅆxyx=ⅆⅆxθxⅆⅆθyθComputeⅆⅆxyxⅆⅆxyx=ⅆⅆθyθx2+1Calculateⅆ2ⅆx2yxwith change of variablesⅆ2ⅆx2yx=ⅆ2ⅆx2θxⅆⅆθyθ+ⅆⅆxθx2ⅆ2ⅆθ2yθCompute second derivativeⅆ2ⅆx2yx=xⅆⅆθyθx2+132+ⅆ2ⅆθ2yθx2+1Apply the change of variables to the ODEx2+1xⅆⅆθyθx2+132+ⅆ2ⅆθ2yθx2+1+xⅆⅆθyθx2+1+yx=0Multiply throughx3ⅆⅆθyθx2+132xⅆⅆθyθx2+132ⅆ2ⅆθ2yθx2x2+1+ⅆ2ⅆθ2yθx2+1+xⅆⅆθyθx2+1+yx=0Simplify ODEⅆ2ⅆθ2yθ+yx=0ODE is that of a harmonic oscillator with given general solutionyθ=_C1+_C2Revert back toxyx=_C1+_C2Use trig identity to simplify=x2+1Simplify solution to the ODEyx=_C1x2+1+_C2x

(4)

ode3x2+1diffyx,x,xxdiffyx,x+4yx=0

ode3x2+1ⅆ2ⅆx2yxxⅆⅆxyx+4yx=0

(5)

ODEStepsode3

Let's solvex2+1ⅆ2ⅆx2yxxⅆⅆxyx+4yx=0Highest derivative means the order of the ODE is2ⅆ2ⅆx2yxIsolate 2nd derivativeⅆ2ⅆx2yx=4yxx21xⅆⅆxyxx21Group terms withyxon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2yx4yxx21+xⅆⅆxyxx21=0Multiply by denominators of ODEx2+1ⅆ2ⅆx2yxxⅆⅆxyx+4yx=0Make a change of variablesθ=arccosxCalculateⅆⅆxyxwith change of variablesⅆⅆxyx=ⅆⅆxθxⅆⅆθyθComputeⅆⅆxyxⅆⅆxyx=ⅆⅆθyθx2+1Calculateⅆ2ⅆx2yxwith change of variablesⅆ2ⅆx2yx=ⅆ2ⅆx2θxⅆⅆθyθ+ⅆⅆxθx2ⅆ2ⅆθ2yθCompute second derivativeⅆ2ⅆx2yx=xⅆⅆθyθx2+132+ⅆ2ⅆθ2yθx2+1Apply the change of variables to the ODEx2+1xⅆⅆθyθx2+132+ⅆ2ⅆθ2yθx2+1+xⅆⅆθyθx2+1+4yx=0Multiply throughx3ⅆⅆθyθx2+132xⅆⅆθyθx2+132ⅆ2ⅆθ2yθx2x2+1+ⅆ2ⅆθ2yθx2+1+xⅆⅆθyθx2+1+4yx=0Simplify ODEⅆ2ⅆθ2yθ+4yx=0ODE is that of a harmonic oscillator with given general solutionyθ=_C1+_C2Revert back toxyx=_C1+_C2Apply double angle identities to solutionyx=_C1+_C2221Use trig identity to simplify sin=x2+1Simplify solution to the ODEyx=_C1xx2+1+_C22x21

(6)

ode4x2+1diffyx,x,xxdiffyx,x+9yx=0

ode4x2+1ⅆ2ⅆx2yxxⅆⅆxyx+9yx=0

(7)

ODEStepsode4

Let's solvex2+1ⅆ2ⅆx2yxxⅆⅆxyx+9yx=0Highest derivative means the order of the ODE is2ⅆ2ⅆx2yxIsolate 2nd derivativeⅆ2ⅆx2yx=9yxx21xⅆⅆxyxx21Group terms withyxon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2yx9yxx21+xⅆⅆxyxx21=0Multiply by denominators of ODEx2+1ⅆ2ⅆx2yxxⅆⅆxyx+9yx=0Make a change of variablesθ=arccosxCalculateⅆⅆxyxwith change of variablesⅆⅆxyx=ⅆⅆxθxⅆⅆθyθComputeⅆⅆxyxⅆⅆxyx=ⅆⅆθyθx2+1Calculateⅆ2ⅆx2yxwith change of variablesⅆ2ⅆx2yx=ⅆ2ⅆx2θxⅆⅆθyθ+ⅆⅆxθx2ⅆ2ⅆθ2yθCompute second derivativeⅆ2ⅆx2yx=xⅆⅆθyθx2+132+ⅆ2ⅆθ2yθx2+1Apply the change of variables to the ODEx2+1xⅆⅆθyθx2+132+ⅆ2ⅆθ2yθx2+1+xⅆⅆθyθx2+1+9yx=0Multiply throughx3ⅆⅆθyθx2+132xⅆⅆθyθx2+132ⅆ2ⅆθ2yθx2x2+1+ⅆ2ⅆθ2yθx2+1+xⅆⅆθyθx2+1+9yx=0Simplify ODEⅆ2ⅆθ2yθ+9yx=0ODE is that of a harmonic oscillator with given general solutionyθ=_C1+_C2Revert back toxyx=_C1+_C2

(8)

ode5x2+1diffyx,x,xxdiffyx,x4yx=0

ode5x2+1ⅆ2ⅆx2yxxⅆⅆxyx4yx=0

(9)

ODEStepsode5

Let's solvex2+1ⅆ2ⅆx2yxxⅆⅆxyx4yx=0Highest derivative means the order of the ODE is2ⅆ2ⅆx2yxIsolate 2nd derivativeⅆ2ⅆx2yx=4yxx21xⅆⅆxyxx21Group terms withyxon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2yx+4yxx21+xⅆⅆxyxx21=0Multiply by denominators of ODEx2+1ⅆ2ⅆx2yxxⅆⅆxyx4yx=0Make a change of variablesθ=arccosxCalculateⅆⅆxyxwith change of variablesⅆⅆxyx=ⅆⅆxθxⅆⅆθyθComputeⅆⅆxyxⅆⅆxyx=ⅆⅆθyθx2+1Calculateⅆ2ⅆx2yxwith change of variablesⅆ2ⅆx2yx=ⅆ2ⅆx2θxⅆⅆθyθ+ⅆⅆxθx2ⅆ2ⅆθ2yθCompute second derivativeⅆ2ⅆx2yx=xⅆⅆθyθx2+132+ⅆ2ⅆθ2yθx2