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












4












$begingroup$


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.
    Mar 18 at 7:37












  • $begingroup$
    You have a type in your M code, you typed G_y for Gy
    $endgroup$
    – Nasser
    Mar 18 at 7:47










  • $begingroup$
    @Nasser corrected
    $endgroup$
    – enea19
    Mar 18 at 8:47
















4












$begingroup$


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.
    Mar 18 at 7:37












  • $begingroup$
    You have a type in your M code, you typed G_y for Gy
    $endgroup$
    – Nasser
    Mar 18 at 7:47










  • $begingroup$
    @Nasser corrected
    $endgroup$
    – enea19
    Mar 18 at 8:47














4












4








4





$begingroup$


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 Mar 18 at 9:56









m_goldberg

87.9k872198




87.9k872198










asked Mar 18 at 7:11









enea19enea19

877




877








  • 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.
    Mar 18 at 7:37












  • $begingroup$
    You have a type in your M code, you typed G_y for Gy
    $endgroup$
    – Nasser
    Mar 18 at 7:47










  • $begingroup$
    @Nasser corrected
    $endgroup$
    – enea19
    Mar 18 at 8:47














  • 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.
    Mar 18 at 7:37












  • $begingroup$
    You have a type in your M code, you typed G_y for Gy
    $endgroup$
    – Nasser
    Mar 18 at 7:47










  • $begingroup$
    @Nasser corrected
    $endgroup$
    – enea19
    Mar 18 at 8:47








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.
Mar 18 at 7:37






$begingroup$
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.
Mar 18 at 7:37














$begingroup$
You have a type in your M code, you typed G_y for Gy
$endgroup$
– Nasser
Mar 18 at 7:47




$begingroup$
You have a type in your M code, you typed G_y for Gy
$endgroup$
– Nasser
Mar 18 at 7:47












$begingroup$
@Nasser corrected
$endgroup$
– enea19
Mar 18 at 8:47




$begingroup$
@Nasser corrected
$endgroup$
– enea19
Mar 18 at 8:47










1 Answer
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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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    $begingroup$

    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











    $endgroup$


















      4












      $begingroup$

      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











      $endgroup$
















        4












        4








        4





        $begingroup$

        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 Mar 18 at 7:57

























        answered Mar 18 at 7:46









        NasserNasser

        58.6k489206




        58.6k489206






























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