/ Unlabelled / Concerning the origin of four terms in DSolve's hyperbolic solution to an ODE

Senin, 18 Maret 2019

Concerning the origin of four terms in DSolve's hyperbolic solution to an ODE





















3
























$begingroup$





Typing



DSolve[{g''[y] - k^2/Gy g[y] == 0}, g[y], y]



where $k$ and $G_y$ are constants gives
$$left{left{g(y)to C_1 expleft(frac{k y}{sqrt{G_y}}right)+C_2 expleft(-frac{k y}{sqrt{G_y}}right)right}right}$$
This anwer can be expressed in terms of hyperbolic functions
$$g(y)=C_1coshleft(frac{k y}{sqrt{G_y}}right)+C_2sinhleft(frac{k y}{sqrt{G_y}}right)$$
where $C_1$ and $C_2$ are not technically the same as the constants for the exponential form but for the purposes of this question it doesn't matter. When I type ExpToTrig[DSolve[{ g''[y] - k^2/Gy g[y] == 0}, g[y], y]] I get
$$left{left{g(y)to C_1 sinh left(frac{k y}{sqrt{G_y}}right)-C_2 sinh left(frac{k y}{sqrt{G_y}}right)+C_1 cosh left(frac{k y}{sqrt{G_y}}right)+C_2 cosh left(frac{k y}{sqrt{G_y}}right)right}right}$$



How come Mathematica creates four terms instead of two? Doesn't this make finding solutions more difficult?










differential-equations calculus-and-analysis














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  • 1









    $begingroup$

    Mathematica does not know that C1 and C2 are arbitrary constants, but you do, and you can write C1*Sinh+C2*Cosh. There is no mistake here.

    $endgroup$

    – Vsevolod A.

    3 hours ago

























  • $begingroup$

    You have a type in your M code, you typed G_y for Gy

    $endgroup$

    – Nasser

    3 hours ago





















  • $begingroup$

    @Nasser corrected

    $endgroup$

    – enea19

    2 hours ago






























3
























$begingroup$





Typing



DSolve[{g''[y] - k^2/Gy g[y] == 0}, g[y], y]



where $k$ and $G_y$ are constants gives
$$left{left{g(y)to C_1 expleft(frac{k y}{sqrt{G_y}}right)+C_2 expleft(-frac{k y}{sqrt{G_y}}right)right}right}$$
This anwer can be expressed in terms of hyperbolic functions
$$g(y)=C_1coshleft(frac{k y}{sqrt{G_y}}right)+C_2sinhleft(frac{k y}{sqrt{G_y}}right)$$
where $C_1$ and $C_2$ are not technically the same as the constants for the exponential form but for the purposes of this question it doesn't matter. When I type ExpToTrig[DSolve[{ g''[y] - k^2/Gy g[y] == 0}, g[y], y]] I get
$$left{left{g(y)to C_1 sinh left(frac{k y}{sqrt{G_y}}right)-C_2 sinh left(frac{k y}{sqrt{G_y}}right)+C_1 cosh left(frac{k y}{sqrt{G_y}}right)+C_2 cosh left(frac{k y}{sqrt{G_y}}right)right}right}$$



How come Mathematica creates four terms instead of two? Doesn't this make finding solutions more difficult?










differential-equations calculus-and-analysis














share|improve this question






















$endgroup$
















  • 1









    $begingroup$

    Mathematica does not know that C1 and C2 are arbitrary constants, but you do, and you can write C1*Sinh+C2*Cosh. There is no mistake here.

    $endgroup$

    – Vsevolod A.

    3 hours ago

























  • $begingroup$

    You have a type in your M code, you typed G_y for Gy

    $endgroup$

    – Nasser

    3 hours ago





















  • $begingroup$

    @Nasser corrected

    $endgroup$

    – enea19

    2 hours ago


























3






















3














3






$begingroup$





Typing



DSolve[{g''[y] - k^2/Gy g[y] == 0}, g[y], y]



where $k$ and $G_y$ are constants gives
$$left{left{g(y)to C_1 expleft(frac{k y}{sqrt{G_y}}right)+C_2 expleft(-frac{k y}{sqrt{G_y}}right)right}right}$$
This anwer can be expressed in terms of hyperbolic functions
$$g(y)=C_1coshleft(frac{k y}{sqrt{G_y}}right)+C_2sinhleft(frac{k y}{sqrt{G_y}}right)$$
where $C_1$ and $C_2$ are not technically the same as the constants for the exponential form but for the purposes of this question it doesn't matter. When I type ExpToTrig[DSolve[{ g''[y] - k^2/Gy g[y] == 0}, g[y], y]] I get
$$left{left{g(y)to C_1 sinh left(frac{k y}{sqrt{G_y}}right)-C_2 sinh left(frac{k y}{sqrt{G_y}}right)+C_1 cosh left(frac{k y}{sqrt{G_y}}right)+C_2 cosh left(frac{k y}{sqrt{G_y}}right)right}right}$$



How come Mathematica creates four terms instead of two? Doesn't this make finding solutions more difficult?










differential-equations calculus-and-analysis














share|improve this question






















$endgroup$







Typing



DSolve[{g''[y] - k^2/Gy g[y] == 0}, g[y], y]



where $k$ and $G_y$ are constants gives
$$left{left{g(y)to C_1 expleft(frac{k y}{sqrt{G_y}}right)+C_2 expleft(-frac{k y}{sqrt{G_y}}right)right}right}$$
This anwer can be expressed in terms of hyperbolic functions
$$g(y)=C_1coshleft(frac{k y}{sqrt{G_y}}right)+C_2sinhleft(frac{k y}{sqrt{G_y}}right)$$
where $C_1$ and $C_2$ are not technically the same as the constants for the exponential form but for the purposes of this question it doesn't matter. When I type ExpToTrig[DSolve[{ g''[y] - k^2/Gy g[y] == 0}, g[y], y]] I get
$$left{left{g(y)to C_1 sinh left(frac{k y}{sqrt{G_y}}right)-C_2 sinh left(frac{k y}{sqrt{G_y}}right)+C_1 cosh left(frac{k y}{sqrt{G_y}}right)+C_2 cosh left(frac{k y}{sqrt{G_y}}right)right}right}$$



How come Mathematica creates four terms instead of two? Doesn't this make finding solutions more difficult?








differential-equations calculus-and-analysis







differential-equations calculus-and-analysis










share|improve this question



























share|improve this question























share|improve this question





share|improve this question










edited 1 hour ago













m_goldberg



87.6k872198







87.6k872198















asked 4 hours ago













enea19enea19



676







676














  • 1









    $begingroup$

    Mathematica does not know that C1 and C2 are arbitrary constants, but you do, and you can write C1*Sinh+C2*Cosh. There is no mistake here.

    $endgroup$

    – Vsevolod A.

    3 hours ago

























  • $begingroup$

    You have a type in your M code, you typed G_y for Gy

    $endgroup$

    – Nasser

    3 hours ago





















  • $begingroup$

    @Nasser corrected

    $endgroup$

    – enea19

    2 hours ago

























  • 1









    $begingroup$

    Mathematica does not know that C1 and C2 are arbitrary constants, but you do, and you can write C1*Sinh+C2*Cosh. There is no mistake here.

    $endgroup$

    – Vsevolod A.

    3 hours ago

























  • $begingroup$

    You have a type in your M code, you typed G_y for Gy

    $endgroup$

    – Nasser

    3 hours ago





















  • $begingroup$

    @Nasser corrected

    $endgroup$

    – enea19

    2 hours ago














1







1







$begingroup$

Mathematica does not know that C1 and C2 are arbitrary constants, but you do, and you can write C1*Sinh+C2*Cosh. There is no mistake here.

$endgroup$

– Vsevolod A.

3 hours ago











$begingroup$

Mathematica does not know that C1 and C2 are arbitrary constants, but you do, and you can write C1*Sinh+C2*Cosh. There is no mistake here.

$endgroup$

– Vsevolod A.

3 hours ago























$begingroup$

You have a type in your M code, you typed G_y for Gy

$endgroup$

– Nasser

3 hours ago







$begingroup$

You have a type in your M code, you typed G_y for Gy

$endgroup$

– Nasser

3 hours ago



















$begingroup$

@Nasser corrected

$endgroup$

– enea19

2 hours ago







$begingroup$

@Nasser corrected

$endgroup$

– enea19

2 hours ago

















1 Answer

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oldest



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$begingroup$



Mathematica always did this, and I do not know of an option or assumption to avoid this. But I always did something like this for this case



ClearAll[g, y, k, Gy];

sol = DSolve[{g''[y] - k^2/Gy g[y] == 0}, g[y], y];

sol = Simplify[ExpToTrig[sol]];

sol /. {C[1] + C[2] -> C[3], C[1] - C[2] -> C[4]}



Mathematica graphics



btw, Maple also does similar thing here. 



ode:=diff(g(y),y$2)-k^2/Gy*g(y)=0;

sol:=dsolve(ode,g(y));

convert(sol,trig);

simplify(%);



Mathematica graphics



Sometimes you have to manipulate the CAS output a little.









share|improve this answer






















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    1 Answer

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    active



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    1 Answer

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    active



    oldest



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    active



    oldest



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    active



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    3
























    $begingroup$



    Mathematica always did this, and I do not know of an option or assumption to avoid this. But I always did something like this for this case



    ClearAll[g, y, k, Gy];

    sol = DSolve[{g''[y] - k^2/Gy g[y] == 0}, g[y], y];

    sol = Simplify[ExpToTrig[sol]];

    sol /. {C[1] + C[2] -> C[3], C[1] - C[2] -> C[4]}



    Mathematica graphics



    btw, Maple also does similar thing here. 



    ode:=diff(g(y),y$2)-k^2/Gy*g(y)=0;

    sol:=dsolve(ode,g(y));

    convert(sol,trig);

    simplify(%);



    Mathematica graphics



    Sometimes you have to manipulate the CAS output a little.









    share|improve this answer






















    $endgroup$


































      3
























      $begingroup$



      Mathematica always did this, and I do not know of an option or assumption to avoid this. But I always did something like this for this case



      ClearAll[g, y, k, Gy];

      sol = DSolve[{g''[y] - k^2/Gy g[y] == 0}, g[y], y];

      sol = Simplify[ExpToTrig[sol]];

      sol /. {C[1] + C[2] -> C[3], C[1] - C[2] -> C[4]}



      Mathematica graphics



      btw, Maple also does similar thing here. 



      ode:=diff(g(y),y$2)-k^2/Gy*g(y)=0;

      sol:=dsolve(ode,g(y));

      convert(sol,trig);

      simplify(%);



      Mathematica graphics



      Sometimes you have to manipulate the CAS output a little.









      share|improve this answer






















      $endgroup$






























        3






















        3














        3






        $begingroup$



        Mathematica always did this, and I do not know of an option or assumption to avoid this. But I always did something like this for this case



        ClearAll[g, y, k, Gy];

        sol = DSolve[{g''[y] - k^2/Gy g[y] == 0}, g[y], y];

        sol = Simplify[ExpToTrig[sol]];

        sol /. {C[1] + C[2] -> C[3], C[1] - C[2] -> C[4]}



        Mathematica graphics



        btw, Maple also does similar thing here. 



        ode:=diff(g(y),y$2)-k^2/Gy*g(y)=0;

        sol:=dsolve(ode,g(y));

        convert(sol,trig);

        simplify(%);



        Mathematica graphics



        Sometimes you have to manipulate the CAS output a little.









        share|improve this answer






















        $endgroup$





        Mathematica always did this, and I do not know of an option or assumption to avoid this. But I always did something like this for this case



        ClearAll[g, y, k, Gy];

        sol = DSolve[{g''[y] - k^2/Gy g[y] == 0}, g[y], y];

        sol = Simplify[ExpToTrig[sol]];

        sol /. {C[1] + C[2] -> C[3], C[1] - C[2] -> C[4]}



        Mathematica graphics



        btw, Maple also does similar thing here. 



        ode:=diff(g(y),y$2)-k^2/Gy*g(y)=0;

        sol:=dsolve(ode,g(y));

        convert(sol,trig);

        simplify(%);



        Mathematica graphics



        Sometimes you have to manipulate the CAS output a little.









        share|improve this answer

























        share|improve this answer



        share|improve this answer










        edited 3 hours ago







































        answered 3 hours ago













        NasserNasser



        58.3k489206







        58.3k489206


















































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            Concerning the origin of four terms in DSolve's hyperbolic solution to an ODE Rating: 4.5 Diposkan Oleh: Admin

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