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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

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with⁡Student:-ODEs:

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ode1≔x2⁢diff⁡y⁡x,x,x+4⁢x⁢diff⁡y⁡x,x+25⁢x2−9⁢y⁡x=0

ode1≔x2⁢ⅆ2ⅆx2y⁡x+4⁢x⁢ⅆⅆxy⁡x+25⁢x2−9⁢y⁡x=0

(1)
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ODESteps⁡ode1

Let's solvex2⁢ⅆ2ⅆx2y⁡x+4⁢x⁢ⅆⅆxy⁡x+25⁢x2−9⁢y⁡x=0•Highest derivative means the order of the ODE is2ⅆ2ⅆx2y⁡x•Isolate 2nd derivativeⅆ2ⅆx2y⁡x=−25⁢x2−9⁢y⁡xx2−4⁢ⅆⅆxy⁡xx•Group terms withy⁡xon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2y⁡x+4⁢ⅆⅆxy⁡xx+25⁢x2−9⁢y⁡xx2=0•Simplify ODEx2⁢ⅆ2ⅆx2y⁡x+25⁢y⁡x⁢x2+4⁢x⁢ⅆⅆxy⁡x−9⁢y⁡x=0•Make a change of variablest=5⁢x•Computeⅆⅆxy⁡xⅆⅆxy⁡x=5⁢ⅆⅆty⁡t•Compute second derivativeⅆ2ⅆx2y⁡x=25⁢ⅆ2ⅆt2y⁡t•Apply change of variables to the ODEt2⁢ⅆ2ⅆt2y⁡t+y⁡t⁢t2+4⁢t⁢ⅆⅆty⁡t−9⁢y⁡t=0•Make a change of variablesy⁡t=u⁡tt32•Computeⅆⅆty⁡tⅆⅆty⁡t=−3⁢u⁡t2⁢t52+ⅆⅆtu⁡tt32•Computeⅆ2ⅆt2y⁡tⅆ2ⅆt2y⁡t=15⁢u⁡t4⁢t72−3⁢ⅆⅆtu⁡tt52+ⅆ2ⅆt2u⁡tt32•Apply change of variables to the ODEu⁡t⁢t2+ⅆ2ⅆt2u⁡t⁢t2+ⅆⅆtu⁡t⁢t−45⁢u⁡t4=0•ODE is now of the Bessel form•Solution to Bessel ODEu⁡t=c__1⁢BesselJ⁡3⁢52,t+c__2⁢BesselY⁡3⁢52,t•Make the change fromy⁡xback toy⁡ty⁡t=c__1⁢BesselJ⁡3⁢52,t+c__2⁢BesselY⁡3⁢52,tt32•Make the change fromtback toxy⁡x=c__1⁢BesselJ⁡3⁢52,5⁢x+c__2⁢BesselY⁡3⁢52,5⁢x⁢525⁢x32

(2)
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ode2≔−x2+1⁢diff⁡y⁡x,x,x−x⁢diff⁡y⁡x,x+y⁡x=0

ode2≔−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x+y⁡x=0

(3)
> 

ODESteps⁡ode2

Let's solve−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x+y⁡x=0•Highest derivative means the order of the ODE is2ⅆ2ⅆx2y⁡x•Isolate 2nd derivativeⅆ2ⅆx2y⁡x=y⁡xx2−1−x⁢ⅆⅆxy⁡xx2−1•Group terms withy⁡xon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2y⁡x+x⁢ⅆⅆxy⁡xx2−1−y⁡xx2−1=0•Multiply by denominators of ODE−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x+y⁡x=0•Make a change of variablesθ=arccos⁡x•Calculateⅆⅆxy⁡xwith change of variablesⅆⅆxy⁡x=ⅆⅆθy⁡θ⁢ⅆⅆxθ⁡x•Compute1stderivativeⅆⅆxy⁡xⅆⅆxy⁡x=−ⅆⅆθy⁡θ−x2+1•Calculateⅆ2ⅆx2y⁡xwith change of variablesⅆ2ⅆx2y⁡x=ⅆ2ⅆθ2y⁡θ⁢ⅆⅆxθ⁡x2+ⅆ2ⅆx2θ⁡x⁢ⅆⅆθy⁡θ•Compute2ndderivativeⅆ2ⅆx2y⁡xⅆ2ⅆx2y⁡x=ⅆ2ⅆθ2y⁡θ−x2+1−x⁢ⅆⅆθy⁡θ−x2+132•Apply the change of variables to the ODE−x2+1⁢ⅆ2ⅆθ2y⁡θ−x2+1−x⁢ⅆⅆθy⁡θ−x2+132+x⁢ⅆⅆθy⁡θ−x2+1+y⁡x=0•Multiply through−ⅆ2ⅆθ2y⁡θ⁢x2−x2+1+ⅆ2ⅆθ2y⁡θ−x2+1+x3⁢ⅆⅆθy⁡θ−x2+132−x⁢ⅆⅆθy⁡θ−x2+132+x⁢ⅆⅆθy⁡θ−x2+1+y⁡x=0•Simplify ODEⅆ2ⅆθ2y⁡θ+y⁡x=0•ODE is that of a harmonic oscillator with given general solutiony⁡θ=c__1⁢sin⁡θ+c__2⁢cos⁡θ•Revert back toxy⁡x=c__1⁢sin⁡arccos⁡x+c__2⁢cos⁡arccos⁡x•Use trig identity to simplifysin⁡arccos⁡xsin⁡arccos⁡x=−x2+1•Simplify solution to the ODEy⁡x=c__1⁢−x2+1+c__2⁢x

(4)
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ode3≔−x2+1⁢diff⁡y⁡x,x,x−x⁢diff⁡y⁡x,x+4⁢y⁡x=0

ode3≔−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x+4⁢y⁡x=0

(5)
> 

ODESteps⁡ode3

Let's solve−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x+4⁢y⁡x=0•Highest derivative means the order of the ODE is2ⅆ2ⅆx2y⁡x•Isolate 2nd derivativeⅆ2ⅆx2y⁡x=4⁢y⁡xx2−1−x⁢ⅆⅆxy⁡xx2−1•Group terms withy⁡xon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2y⁡x+x⁢ⅆⅆxy⁡xx2−1−4⁢y⁡xx2−1=0•Multiply by denominators of ODE−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x+4⁢y⁡x=0•Make a change of variablesθ=arccos⁡x•Calculateⅆⅆxy⁡xwith change of variablesⅆⅆxy⁡x=ⅆⅆθy⁡θ⁢ⅆⅆxθ⁡x•Compute1stderivativeⅆⅆxy⁡xⅆⅆxy⁡x=−ⅆⅆθy⁡θ−x2+1•Calculateⅆ2ⅆx2y⁡xwith change of variablesⅆ2ⅆx2y⁡x=ⅆ2ⅆθ2y⁡θ⁢ⅆⅆxθ⁡x2+ⅆ2ⅆx2θ⁡x⁢ⅆⅆθy⁡θ•Compute2ndderivativeⅆ2ⅆx2y⁡xⅆ2ⅆx2y⁡x=ⅆ2ⅆθ2y⁡θ−x2+1−x⁢ⅆⅆθy⁡θ−x2+132•Apply the change of variables to the ODE−x2+1⁢ⅆ2ⅆθ2y⁡θ−x2+1−x⁢ⅆⅆθy⁡θ−x2+132+x⁢ⅆⅆθy⁡θ−x2+1+4⁢y⁡x=0•Multiply through−ⅆ2ⅆθ2y⁡θ⁢x2−x2+1+ⅆ2ⅆθ2y⁡θ−x2+1+x3⁢ⅆⅆθy⁡θ−x2+132−x⁢ⅆⅆθy⁡θ−x2+132+x⁢ⅆⅆθy⁡θ−x2+1+4⁢y⁡x=0•Simplify ODEⅆ2ⅆθ2y⁡θ+4⁢y⁡x=0•ODE is that of a harmonic oscillator with given general solutiony⁡θ=c__1⁢sin⁡2⁢θ+c__2⁢cos⁡2⁢θ•Revert back toxy⁡x=c__1⁢sin⁡2⁢arccos⁡x+c__2⁢cos⁡2⁢arccos⁡x•Apply double angle identities to solutiony⁡x=c__1⁢sin⁡arccos⁡x⁢cos⁡arccos⁡x+c__2⁢2⁢cos⁡arccos⁡x2−1•Use trig identity to simplify sinsin⁡arccos⁡x=−x2+1•Simplify solution to the ODEy⁡x=c__1⁢x⁢−x2+1+c__2⁢2⁢x2−1

(6)
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ode4≔−x2+1⁢diff⁡y⁡x,x,x−x⁢diff⁡y⁡x,x+9⁢y⁡x=0

ode4≔−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x+9⁢y⁡x=0

(7)
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ODESteps⁡ode4

Let's solve−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x+9⁢y⁡x=0•Highest derivative means the order of the ODE is2ⅆ2ⅆx2y⁡x•Isolate 2nd derivativeⅆ2ⅆx2y⁡x=9⁢y⁡xx2−1−x⁢ⅆⅆxy⁡xx2−1•Group terms withy⁡xon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2y⁡x+x⁢ⅆⅆxy⁡xx2−1−9⁢y⁡xx2−1=0•Multiply by denominators of ODE−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x+9⁢y⁡x=0•Make a change of variablesθ=arccos⁡x•Calculateⅆⅆxy⁡xwith change of variablesⅆⅆxy⁡x=ⅆⅆθy⁡θ⁢ⅆⅆxθ⁡x•Compute1stderivativeⅆⅆxy⁡xⅆⅆxy⁡x=−ⅆⅆθy⁡θ−x2+1•Calculateⅆ2ⅆx2y⁡xwith change of variablesⅆ2ⅆx2y⁡x=ⅆ2ⅆθ2y⁡θ⁢ⅆⅆxθ⁡x2+ⅆ2ⅆx2θ⁡x⁢ⅆⅆθy⁡θ•Compute2ndderivativeⅆ2ⅆx2y⁡xⅆ2ⅆx2y⁡x=ⅆ2ⅆθ2y⁡θ−x2+1−x⁢ⅆⅆθy⁡θ−x2+132•Apply the change of variables to the ODE−x2+1⁢ⅆ2ⅆθ2y⁡θ−x2+1−x⁢ⅆⅆθy⁡θ−x2+132+x⁢ⅆⅆθy⁡θ−x2+1+9⁢y⁡x=0•Multiply through−ⅆ2ⅆθ2y⁡θ⁢x2−x2+1+ⅆ2ⅆθ2y⁡θ−x2+1+x3⁢ⅆⅆθy⁡θ−x2+132−x⁢ⅆⅆθy⁡θ−x2+132+x⁢ⅆⅆθy⁡θ−x2+1+9⁢y⁡x=0•Simplify ODEⅆ2ⅆθ2y⁡θ+9⁢y⁡x=0•ODE is that of a harmonic oscillator with given general solutiony⁡θ=c__1⁢sin⁡3⁢θ+c__2⁢cos⁡3⁢θ•Revert back toxy⁡x=c__1⁢sin⁡3⁢arccos⁡x+c__2⁢cos⁡3⁢arccos⁡x

(8)
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ode5≔−x2+1⁢diff⁡y⁡x,x,x−x⁢diff⁡y⁡x,x−4⁢y⁡x=0

ode5≔−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x−4⁢y⁡x=0

(9)
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ODESteps⁡ode5

Let's solve−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x−4⁢y⁡x=0•Highest derivative means the order of the ODE is2ⅆ2ⅆx2y⁡x•Isolate 2nd derivativeⅆ2ⅆx2y⁡x=−4⁢y⁡xx2−1−x⁢ⅆⅆxy⁡xx2−1•Group terms withy⁡xon the lhs of the ODE and the rest on the rhs of the ODE; ODE is linearⅆ2ⅆx2y⁡x+x⁢ⅆⅆxy⁡xx2−1+4⁢y⁡xx2−1=0•Multiply by denominators of ODE−x2+1⁢ⅆ2ⅆx2y⁡x−x⁢ⅆⅆxy⁡x−4⁢y⁡x=0•Make a change of variablesθ=arccos⁡x•Calculateⅆⅆxy⁡xwith change of variablesⅆⅆxy⁡x=ⅆⅆθy⁡θ⁢ⅆⅆxθ⁡x•Compute1stderivativeⅆⅆxy⁡xⅆⅆxy⁡x=−ⅆⅆθy⁡θ−x2+1•Calculateⅆ2ⅆx2y⁡xwith change of variablesⅆ2ⅆx2y⁡x=ⅆ2ⅆθ2y⁡θ⁢ⅆⅆxθ⁡x2+ⅆ2ⅆx2θ⁡x⁢ⅆⅆθy⁡θ•Compute2ndderivativeⅆ2ⅆx2y⁡xⅆ2ⅆx2y⁡x=ⅆ2ⅆθ2y⁡θ−x2+1−x⁢ⅆⅆθy⁡θ−x2+132•Apply the change of variables to the ODE−x2+1⁢ⅆ2ⅆθ2y⁡θ−x2+1−x⁢ⅆⅆθy⁡θ−x2+132+x⁢ⅆⅆθy⁡θ−x2+1−4⁢y⁡x=0•Multiply through−ⅆ2ⅆθ2y⁡θ⁢x2−x2+1+ⅆ2ⅆθ2y⁡θ−x2+1+x3⁢ⅆⅆθy⁡θ−x2+132−x⁢ⅆⅆθy⁡θ−x2+132+x⁢ⅆⅆθy⁡θ−x2+1−4⁢y⁡x=0•Simplify ODEⅆ2ⅆθ2y⁡θ−4⁢y⁡x=0•ODE is second order linear with characteristic polynomial that is the difference of squares with given general solutiony⁡θ=c__1⁢ⅇ2⁢θ+c__2⁢ⅇ−2⁢θ•Revert back toxy⁡x=c__1⁢ⅇ2⁢arccos⁡x+c__2⁢ⅇ−2⁢arccos⁡x

(10)

See Also

diff

Int

Student

Student[ODEs]

Student[ODEs][ODESteps]