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Student[ODEs]

  

ODESteps

  

Show a step-by-step solution process for ODEs, IVPs, or systems

 

Calling Sequence

Parameters

Description

Examples

Calling Sequence

ODESteps(ODE)

ODESteps(ODE, y(x))

ODESteps(sys)

Parameters

ODE

-

an ordinary differential equation

y

-

name ; the dependent variable

x

-

name ; the independent variable

sys

-

set ; an ODE system including initial values

Description

• 

The ODESteps() command solves an ordinary differential equation (ODE) or system of ODEs.

• 

The input may include a corresponding set of initial values, which would make it an initial value problem (IVP).

• 

The output shows a series of steps in the solving process.

• 

The following types of ODEs and ODE systems and/or solving methods are considered:

First Order ODEs

First Order IVPs

Second Order ODEs

Second Order IVPs

Cauchy-Euler Equations

Series Solutions

Special Function Solutions

 

Systems of ODEs

Systems of ODEs with IVP

 

 

Examples

withStudentODEs:

A first order ODE:

ode1t2zt+1+zt2t1diffzt,t=0

ode1t2zt+1+zt2t1ⅆⅆtzt=0

(1)

ODEStepsode1

Let's solvet2zt+1+zt2t1ⅆⅆtzt=0Highest derivative means the order of the ODE is1ⅆⅆtztSeparate variablesⅆⅆtztzt2zt+1=t2t1Integrate both sides with respect totⅆⅆtztzt2zt+1ⅆt=t2t1ⅆt+_C1Evaluate integralzt22zt+lnzt+1=t22tlnt1+_C1

(2)

A first order IVP:

ivp1t2zt+1+zt2t1diffzt,t=0,z3=1

ivp1t2zt+1+zt2t1ⅆⅆtzt=0,z3=1

(3)

ODEStepsivp1

Let's solvet2zt+1+zt2t1ⅆⅆtzt=0,z3=1Highest derivative means the order of the ODE is1ⅆⅆtztSeparate variablesⅆⅆtztzt2zt+1=t2t1Integrate both sides with respect totⅆⅆtztzt2zt+1ⅆt=t2t1ⅆt+_C1Evaluate integralzt22zt+lnzt+1=t22tlnt1+_C1Use initial conditionz3=112+ln2=152ln2+_C1Solve for_C1_C1=7+2ln2Solution to the IVPzt22zt+lnzt+1=t22tlnt1+7+2ln2

(4)

A second order ODE:

ode22xdiffyx,x9x2+2diffyx,x+x2+1diffyx,x,x=0

ode22xⅆⅆxyx9x2+2ⅆⅆxyx+x2+1ⅆ2ⅆx2yx=0

(5)

ODEStepsode2

Let's solve2xⅆⅆxyx9x2+2ⅆⅆxyx+x2+1ⅆ2ⅆx2yx=0Highest derivative means the order of the ODE is2ⅆ2ⅆx2yxMake substitutionu=ⅆⅆxyxto reduce order of ODE2xux9x2+2ux+x2+1ⅆⅆxux=0Check if ODE is exactODE is exact if the lhs is the total derivative of aC2function=0Compute derivative of lhsxFx,u+uFx,uⅆⅆxux=0Evaluate derivatives2x=2xCondition met, ODE is exactExact ODE implies solution will be of this formFx,u=_C1,Mx,u=xFx,u,Nx,u=uFx,uSolve forFx,uby integratingMx,uwith respect toxFx,u=+_F1uEvaluate integralFx,u=x2u3x3+_F1uTake derivative ofFx,uwith respect touNx,u=uFx,uCompute derivativex2+2u+1=x2+ⅆⅆu_F1uIsolate forⅆⅆu_F1uⅆⅆu_F1u=2u+1Solve for_F1u_F1u=u2+uSubstitute_F1uinto equation forFx,uFx,u=x2u3x3+u2+uSubstituteFx,uinto the solution of the ODEx2u3x3+u2+u=_C1Solve foruxux=x2212x4+12x3+2x2+4_C1+12,ux=x2212+x4+12x3+2x2+4_C1+12Solve 1st ODE foruxux=x2212x4+12x3+2x2+4_C1+12Make substitutionu=ⅆⅆxyxⅆⅆxyx=x2212x4+12x3+2x2+4_C1+12Integrate both sides to solve foryxⅆⅆxyxⅆx=x2212x4+12x3+2x2+4_C1+12ⅆx+_C2Compute lhsyx=x2212x4+12x3+2x2+4_C1+12ⅆx+_C2