Concerning the origin of four terms in DSolve's hyperbolic solution to an ODE The Next CEO of Stack OverflowSymbolic integration of a indexed product term?What should I learn from DSolve working better with a named constant than a number in this case?Solution of Coupled second-order ODEs and plot the diagramWhy does the Part::partw error appear when using DSolve?DSolve returns a Solve expressionHow to use DEigensystem with periodic boundary conditions on the derivative?Why does the DSolve solution to the Cauchy-Euler differential equation not match the analytical solution?why Mathematica gives zero as eigenvalue for $y''+lambda y=0$ for these boundary conditions?Oversimplify of products of square roots in DSolve solutionExtra constant in solution of ODE using DSolve
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Concerning the origin of four terms in DSolve's hyperbolic solution to an ODE
The Next CEO of Stack OverflowSymbolic integration of a indexed product term?What should I learn from DSolve working better with a named constant than a number in this case?Solution of Coupled second-order ODEs and plot the diagramWhy does the Part::partw error appear when using DSolve?DSolve returns a Solve expressionHow to use DEigensystem with periodic boundary conditions on the derivative?Why does the DSolve solution to the Cauchy-Euler differential equation not match the analytical solution?why Mathematica gives zero as eigenvalue for $y''+lambda y=0$ for these boundary conditions?Oversimplify of products of square roots in DSolve solutionExtra constant in solution of ODE using DSolve
$begingroup$
Typing
DSolve[g''[y] - k^2/Gy g[y] == 0, g[y], y]
where $k$ and $G_y$ are constants gives
$$leftleftg(y)to C_1 expleft(frack ysqrtG_yright)+C_2 expleft(-frack ysqrtG_yright)rightright$$
This anwer can be expressed in terms of hyperbolic functions
$$g(y)=C_1coshleft(frack ysqrtG_yright)+C_2sinhleft(frack ysqrtG_yright)$$
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
$$leftleftg(y)to C_1 sinh left(frack ysqrtG_yright)-C_2 sinh left(frack ysqrtG_yright)+C_1 cosh left(frack ysqrtG_yright)+C_2 cosh left(frack ysqrtG_yright)rightright$$
How come Mathematica creates four terms instead of two? Doesn't this make finding solutions more difficult?
differential-equations calculus-and-analysis
edited Mar 18 at 9:56
m_goldberg
87.9k872198
asked Mar 18 at 7:11
enea19enea19
877
$endgroup$
add a comment |
$begingroup$
Typing
DSolve[g''[y] - k^2/Gy g[y] == 0, g[y], y]
where $k$ and $G_y$ are constants gives
$$leftleftg(y)to C_1 expleft(frack ysqrtG_yright)+C_2 expleft(-frack ysqrtG_yright)rightright$$
This anwer can be expressed in terms of hyperbolic functions
$$g(y)=C_1coshleft(frack ysqrtG_yright)+C_2sinhleft(frack ysqrtG_yright)$$
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
$$leftleftg(y)to C_1 sinh left(frack ysqrtG_yright)-C_2 sinh left(frack ysqrtG_yright)+C_1 cosh left(frack ysqrtG_yright)+C_2 cosh left(frack ysqrtG_yright)rightright$$
How come Mathematica creates four terms instead of two? Doesn't this make finding solutions more difficult?
differential-equations calculus-and-analysis
edited Mar 18 at 9:56
m_goldberg
87.9k872198
asked Mar 18 at 7:11
enea19enea19
877
$endgroup$
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 typedG_yforGy
$endgroup$
– Nasser
Mar 18 at 7:47
$begingroup$
@Nasser corrected
$endgroup$
– enea19
Mar 18 at 8:47
add a comment |
$begingroup$
Typing
DSolve[g''[y] - k^2/Gy g[y] == 0, g[y], y]
where $k$ and $G_y$ are constants gives
$$leftleftg(y)to C_1 expleft(frack ysqrtG_yright)+C_2 expleft(-frack ysqrtG_yright)rightright$$
This anwer can be expressed in terms of hyperbolic functions
$$g(y)=C_1coshleft(frack ysqrtG_yright)+C_2sinhleft(frack ysqrtG_yright)$$
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
$$leftleftg(y)to C_1 sinh left(frack ysqrtG_yright)-C_2 sinh left(frack ysqrtG_yright)+C_1 cosh left(frack ysqrtG_yright)+C_2 cosh left(frack ysqrtG_yright)rightright$$
How come Mathematica creates four terms instead of two? Doesn't this make finding solutions more difficult?
differential-equations calculus-and-analysis
edited Mar 18 at 9:56
m_goldberg
87.9k872198
asked Mar 18 at 7:11
enea19enea19
877
$endgroup$
Typing
DSolve[g''[y] - k^2/Gy g[y] == 0, g[y], y]
where $k$ and $G_y$ are constants gives
$$leftleftg(y)to C_1 expleft(frack ysqrtG_yright)+C_2 expleft(-frack ysqrtG_yright)rightright$$
This anwer can be expressed in terms of hyperbolic functions
$$g(y)=C_1coshleft(frack ysqrtG_yright)+C_2sinhleft(frack ysqrtG_yright)$$
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
$$leftleftg(y)to C_1 sinh left(frack ysqrtG_yright)-C_2 sinh left(frack ysqrtG_yright)+C_1 cosh left(frack ysqrtG_yright)+C_2 cosh left(frack ysqrtG_yright)rightright$$
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
edited Mar 18 at 9:56
m_goldberg
87.9k872198
asked Mar 18 at 7:11
enea19enea19
877
edited Mar 18 at 9:56
m_goldberg
87.9k872198
asked Mar 18 at 7:11
enea19enea19
877
edited Mar 18 at 9:56
m_goldberg
87.9k872198
edited Mar 18 at 9:56
m_goldberg
87.9k872198
edited Mar 18 at 9:56
m_goldberg
87.9k872198
87.9k872198
asked Mar 18 at 7:11
enea19enea19
877
asked Mar 18 at 7:11
enea19enea19
877
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 typedG_yforGy
$endgroup$
– Nasser
Mar 18 at 7:47
$begingroup$
@Nasser corrected
$endgroup$
– enea19
Mar 18 at 8:47
add a comment |
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 typedG_yforGy
$endgroup$
– 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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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]
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(%);
Sometimes you have to manipulate the CAS output a little.
answered Mar 18 at 7:46
NasserNasser
58.6k489206
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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]
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(%);
Sometimes you have to manipulate the CAS output a little.
answered Mar 18 at 7:46
NasserNasser
58.6k489206
$endgroup$
add a comment |
$begingroup$
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]
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(%);
Sometimes you have to manipulate the CAS output a little.
answered Mar 18 at 7:46
NasserNasser
58.6k489206
$endgroup$
add a comment |
$begingroup$
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]
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(%);
Sometimes you have to manipulate the CAS output a little.
answered Mar 18 at 7:46
NasserNasser
58.6k489206
$endgroup$
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]
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(%);
Sometimes you have to manipulate the CAS output a little.
answered Mar 18 at 7:46
NasserNasser
58.6k489206
edited Mar 18 at 7:57
answered Mar 18 at 7:46
NasserNasser
58.6k489206
answered Mar 18 at 7:46
NasserNasser
58.6k489206
answered Mar 18 at 7:46
NasserNasser
58.6k489206
58.6k489206
add a comment |
add a comment |
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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.
Mar 18 at 7:37
$begingroup$
You have a type in your M code, you typed
G_yforGy$endgroup$
– Nasser
Mar 18 at 7:47
$begingroup$
@Nasser corrected
$endgroup$
– enea19
Mar 18 at 8:47