First Order IVPs - Maple Help
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ODE Steps for First Order IVPs

 

Overview

Examples

Overview

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This help page gives a few examples of using the command ODESteps to solve first order initial value problems.

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

Examples

withStudent:-ODEs:

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

ivp1t2zt+1+zt2t1ⅆⅆtzt=0,z3=1

(1)

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_C1C1=7+2ln2Substitute_C1=7+2ln2into general solution and simplifyzt22zt+lnzt+1=t22tlnt1+7+2ln2Solution to the IVPzt22zt+lnzt+1=t22tlnt1+7+2ln2

(2)

ivp22xyx9x2+2yx+x2+1diffyx,x=0,y0=1

ivp22xyx9x2+2yx+x2+1ⅆⅆxyx=0,y0=1

(3)

ODEStepsivp2

Let's solve2xyx9x2+2yx+x2+1ⅆⅆxyx=0,y0=1Highest derivative means the order of the ODE is1ⅆⅆxyxCheck if ODE is exactODE is exact if the lhs is the total derivative of aC2functionⅆⅆxFx,yx=0Compute derivative of lhsxFx,y+yFx,yⅆⅆxyx=0Evaluate derivatives2x=2xCondition met, ODE is exactExact ODE implies solution will be of this formFx,y=C1,Mx,y=xFx,y,Nx,y=yFx,ySolve forFx,yby integratingMx,ywith respect toxFx,y=9x2+2xyⅆx+_F1yEvaluate integralFx,y=3x3+x2y+_F1yTake derivative ofFx,ywith respect toyNx,y=yFx,yCompute derivativex2+2y+1=x2+ⅆⅆy_F1yIsolate forⅆⅆy_F1yⅆⅆy_F1y=2y+1Solve for_F1y_F1y=y2+ySubstitute_F1yinto equation forFx,yFx,y=3x3+x2y+y2+ySubstituteFx,yinto the solution of the ODE3x3+x2y+y2+y=C1Solve foryxyx=x2212x4+12x3+2x2+4C1+12,yx=x2212+x4+12x3+2x2+4C1+12Use initial conditiony0=11=124C1+12Solution does not satisfy initial conditionUse initial conditiony0=11=12+4C1+12Solve for_C1C1=2Substitute_C1=2into general solution and simplifyyx=x2212+x4+12x3+2x2+92Solution to the IVPyx=x2212+x4+12x3+2x2+92

(4)

ivp3diffyx,xyxxexpx=0,ya=b

ivp3ⅆⅆxyxyxxⅇx=0,ya=b

(5)

ODEStepsivp3

Let's solveⅆⅆxyxyxxⅇx=0,ya=bHighest derivative means the order of the ODE is1ⅆⅆxyxIsolate the derivativeⅆⅆxyx=yx+xⅇxGroup terms withyxon the lhs of the ODE and the rest on the rhs of the ODEⅆⅆxyxyx=xⅇxThe ODE is linear; multiply by an integrating factorμxμxⅆⅆxyxyx=μxxⅇxAssume the lhs of the ODE is the total derivativeⅆⅆxμxyxμxⅆⅆxyxyx=ⅆⅆxμxyx+μxⅆⅆxyxIsolateⅆⅆxμxⅆⅆxμx=μxSolve to find the integrating factorμx=ⅇxIntegrate both sides with respect toxⅆⅆxμxyxⅆx=μxxⅇxⅆx+C1Evaluate the integral on the lhsμxyx=μxxⅇxⅆx+C1Solve foryxyx=μxxⅇxⅆx+C1μxSubstituteμx=ⅇxyx=ⅇxxⅇxⅆx+C1ⅇxEvaluate the integrals on the rhsyx=x22+C1ⅇxSimplifyyx=ⅇxx2+2C12Use initial conditionya=bb=ⅇaa2+2C12Solve for_C1C1=ⅇaa22b2ⅇaSubstitute_C1=ⅇaa22b2ⅇainto general solution and simplifyyx=ⅇxa2+2ⅇab+x22Solution to the IVPyx=ⅇxa2+2ⅇab+x22

(6)

See Also

diff

Int

Student

Student[ODEs]

Student[ODEs][ODESteps]

 


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