Why is DSolve different when the equation is built programmatically?
4
$begingroup$
Here is a simple differential equation.
mwe=y[x]/.First@DSolve [{y'[x]==x-2*x*y[x],y[0]==0},y[x],x]
I get the answer $frac{1}{2}-frac{e^{-x^2}}{2}$
If I build up the same equation in a simple program, I get a different answer:
lsolve[r_,q_,a_,eta_]=y[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x];
nwe=lsolve[x,2*x,0,0]
After simplifying, I get the answer $frac{1}{2}-frac{e^{-2x^2}}{2}$. Where did the extra factor of $2$ in the exponent come from? And how can I make it go away?
differential-equations variable
edited Feb 12 at 20:32
Peter Mortensen
33627
asked Feb 12 at 15:25
Joe CorneliJoe Corneli
1335
$endgroup$
add a comment |
4
$begingroup$
Here is a simple differential equation.
mwe=y[x]/.First@DSolve [{y'[x]==x-2*x*y[x],y[0]==0},y[x],x]
I get the answer $frac{1}{2}-frac{e^{-x^2}}{2}$
If I build up the same equation in a simple program, I get a different answer:
lsolve[r_,q_,a_,eta_]=y[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x];
nwe=lsolve[x,2*x,0,0]
After simplifying, I get the answer $frac{1}{2}-frac{e^{-2x^2}}{2}$. Where did the extra factor of $2$ in the exponent come from? And how can I make it go away?
differential-equations variable
edited Feb 12 at 20:32
Peter Mortensen
33627
asked Feb 12 at 15:25
Joe CorneliJoe Corneli
1335
$endgroup$
$begingroup$
Here are the docs on this issue, in case it helps someone else. reference.wolfram.com/language/tutorial/…
$endgroup$
– Joe Corneli
Feb 12 at 15:42
add a comment |
4
4
4
$begingroup$
Here is a simple differential equation.
mwe=y[x]/.First@DSolve [{y'[x]==x-2*x*y[x],y[0]==0},y[x],x]
I get the answer $frac{1}{2}-frac{e^{-x^2}}{2}$
If I build up the same equation in a simple program, I get a different answer:
lsolve[r_,q_,a_,eta_]=y[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x];
nwe=lsolve[x,2*x,0,0]
After simplifying, I get the answer $frac{1}{2}-frac{e^{-2x^2}}{2}$. Where did the extra factor of $2$ in the exponent come from? And how can I make it go away?
differential-equations variable
edited Feb 12 at 20:32
Peter Mortensen
33627
asked Feb 12 at 15:25
Joe CorneliJoe Corneli
1335
$endgroup$
Here is a simple differential equation.
mwe=y[x]/.First@DSolve [{y'[x]==x-2*x*y[x],y[0]==0},y[x],x]
I get the answer $frac{1}{2}-frac{e^{-x^2}}{2}$
If I build up the same equation in a simple program, I get a different answer:
lsolve[r_,q_,a_,eta_]=y[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x];
nwe=lsolve[x,2*x,0,0]
After simplifying, I get the answer $frac{1}{2}-frac{e^{-2x^2}}{2}$. Where did the extra factor of $2$ in the exponent come from? And how can I make it go away?
differential-equations variable
differential-equations variable
edited Feb 12 at 20:32
Peter Mortensen
33627
asked Feb 12 at 15:25
Joe CorneliJoe Corneli
1335
edited Feb 12 at 20:32
Peter Mortensen
33627
asked Feb 12 at 15:25
Joe CorneliJoe Corneli
1335
edited Feb 12 at 20:32
Peter Mortensen
33627
edited Feb 12 at 20:32
Peter Mortensen
33627
edited Feb 12 at 20:32
Peter Mortensen
33627
33627
asked Feb 12 at 15:25
Joe CorneliJoe Corneli
1335
asked Feb 12 at 15:25
Joe CorneliJoe Corneli
1335
asked Feb 12 at 15:25
Joe CorneliJoe Corneli
1335
1335
$begingroup$
Here are the docs on this issue, in case it helps someone else. reference.wolfram.com/language/tutorial/…
$endgroup$
– Joe Corneli
Feb 12 at 15:42
add a comment |
$begingroup$
Here are the docs on this issue, in case it helps someone else. reference.wolfram.com/language/tutorial/…
$endgroup$
– Joe Corneli
Feb 12 at 15:42
$begingroup$
Here are the docs on this issue, in case it helps someone else. reference.wolfram.com/language/tutorial/…
$endgroup$
– Joe Corneli
Feb 12 at 15:42
$begingroup$
Here are the docs on this issue, in case it helps someone else. reference.wolfram.com/language/tutorial/…
$endgroup$
– Joe Corneli
Feb 12 at 15:42
add a comment |
1 Answer
1
active
oldest
votes
9
$begingroup$
You attempt to solve the ODE at the time of definition of lsolve. At this point, q does not depend on x. You really want to use SetDelayed here:
lsolve[r_, q_, a_, eta_] :=
y[x] /. First@DSolve[{y'[x] == r - q*y[x], y[a] == eta}, y[x], x];
nwe = lsolve[x, 2*x, 0, 0] // Simplify
1/2 - E^-x^2/2
answered Feb 12 at 15:31
Henrik SchumacherHenrik Schumacher
60.3k583169
$endgroup$
$begingroup$
Thanks, I'm glad I asked. Even though my version was syntactically correct, it would have been nice if Mathematica produced a warning!
$endgroup$
– Joe Corneli
Feb 12 at 15:41
$begingroup$
How should Mathematica be able to detect that this does not align with your expectations? Both syntaxesf[x_] = ...andf[x_] := ...have the applications, e.g., the former can be considered as a shorthand forf[x_] := Evaluate[...]. This is usefull whenever a symbolic compuation can already performed at the time of definition so that it need not be redone every timefis called.
$endgroup$
– Henrik Schumacher
Feb 12 at 17:17
$begingroup$
As you said in the answer: "At this point,qdoes not depend onx." I think Mathematica should be able to tell that I was building an expression that does depend onxand then generate a warning (but certainly not an error) very much along the lines of what you told me. Otherwise everyone who doesn't know about:=will end up asking a question similar to mine. How to preempt that and automate your answer? :-)
$endgroup$
– Joe Corneli
Feb 13 at 13:18
$begingroup$
When an assignment is made with=,Mathematicacan only set the left hand side to the value of the right hand side at the time of assignment. Evaluatey[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x](just in case: in a fresh kernel) and you will see that the result is(E^(-q x) (E^(a q) eta q - E^(a q) r + E^(q x) r))/q. Because Mathematica solves the ODE at this point in time and becauseqis independent ofx. Of course, you may tell Mathematica thatqshould depend onxby usingq[x]instead ofq, but I don't think you will like the result either.
$endgroup$
– Henrik Schumacher
Feb 13 at 13:24
$begingroup$
I don't understand what you mean by "automation". The difference between=and:=lies in the time of execution of the right hand side. If you want that the ODE is solved only after a precise value ofqis supplied, you have to use:=. And yes, there are a lot of questions on the subtle differences of=and:=on this site. And I have to admit that was also something that was quite miraculous to me when I started to use the language. It helps a bit to recall that Mathematica, in its core, is nothing else but a term rewriting system.
$endgroup$
– Henrik Schumacher
Feb 13 at 13:28
|
show 3 more comments
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1 Answer
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1
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9
$begingroup$
You attempt to solve the ODE at the time of definition of lsolve. At this point, q does not depend on x. You really want to use SetDelayed here:
lsolve[r_, q_, a_, eta_] :=
y[x] /. First@DSolve[{y'[x] == r - q*y[x], y[a] == eta}, y[x], x];
nwe = lsolve[x, 2*x, 0, 0] // Simplify
1/2 - E^-x^2/2
answered Feb 12 at 15:31
Henrik SchumacherHenrik Schumacher
60.3k583169
$endgroup$
$begingroup$
Thanks, I'm glad I asked. Even though my version was syntactically correct, it would have been nice if Mathematica produced a warning!
$endgroup$
– Joe Corneli
Feb 12 at 15:41
$begingroup$
How should Mathematica be able to detect that this does not align with your expectations? Both syntaxesf[x_] = ...andf[x_] := ...have the applications, e.g., the former can be considered as a shorthand forf[x_] := Evaluate[...]. This is usefull whenever a symbolic compuation can already performed at the time of definition so that it need not be redone every timefis called.
$endgroup$
– Henrik Schumacher
Feb 12 at 17:17
$begingroup$
As you said in the answer: "At this point,qdoes not depend onx." I think Mathematica should be able to tell that I was building an expression that does depend onxand then generate a warning (but certainly not an error) very much along the lines of what you told me. Otherwise everyone who doesn't know about:=will end up asking a question similar to mine. How to preempt that and automate your answer? :-)
$endgroup$
– Joe Corneli
Feb 13 at 13:18
$begingroup$
When an assignment is made with=,Mathematicacan only set the left hand side to the value of the right hand side at the time of assignment. Evaluatey[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x](just in case: in a fresh kernel) and you will see that the result is(E^(-q x) (E^(a q) eta q - E^(a q) r + E^(q x) r))/q. Because Mathematica solves the ODE at this point in time and becauseqis independent ofx. Of course, you may tell Mathematica thatqshould depend onxby usingq[x]instead ofq, but I don't think you will like the result either.
$endgroup$
– Henrik Schumacher
Feb 13 at 13:24
$begingroup$
I don't understand what you mean by "automation". The difference between=and:=lies in the time of execution of the right hand side. If you want that the ODE is solved only after a precise value ofqis supplied, you have to use:=. And yes, there are a lot of questions on the subtle differences of=and:=on this site. And I have to admit that was also something that was quite miraculous to me when I started to use the language. It helps a bit to recall that Mathematica, in its core, is nothing else but a term rewriting system.
$endgroup$
– Henrik Schumacher
Feb 13 at 13:28
|
show 3 more comments
9
$begingroup$
You attempt to solve the ODE at the time of definition of lsolve. At this point, q does not depend on x. You really want to use SetDelayed here:
lsolve[r_, q_, a_, eta_] :=
y[x] /. First@DSolve[{y'[x] == r - q*y[x], y[a] == eta}, y[x], x];
nwe = lsolve[x, 2*x, 0, 0] // Simplify
1/2 - E^-x^2/2
answered Feb 12 at 15:31
Henrik SchumacherHenrik Schumacher
60.3k583169
$endgroup$
$begingroup$
Thanks, I'm glad I asked. Even though my version was syntactically correct, it would have been nice if Mathematica produced a warning!
$endgroup$
– Joe Corneli
Feb 12 at 15:41
$begingroup$
How should Mathematica be able to detect that this does not align with your expectations? Both syntaxesf[x_] = ...andf[x_] := ...have the applications, e.g., the former can be considered as a shorthand forf[x_] := Evaluate[...]. This is usefull whenever a symbolic compuation can already performed at the time of definition so that it need not be redone every timefis called.
$endgroup$
– Henrik Schumacher
Feb 12 at 17:17
$begingroup$
As you said in the answer: "At this point,qdoes not depend onx." I think Mathematica should be able to tell that I was building an expression that does depend onxand then generate a warning (but certainly not an error) very much along the lines of what you told me. Otherwise everyone who doesn't know about:=will end up asking a question similar to mine. How to preempt that and automate your answer? :-)
$endgroup$
– Joe Corneli
Feb 13 at 13:18
$begingroup$
When an assignment is made with=,Mathematicacan only set the left hand side to the value of the right hand side at the time of assignment. Evaluatey[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x](just in case: in a fresh kernel) and you will see that the result is(E^(-q x) (E^(a q) eta q - E^(a q) r + E^(q x) r))/q. Because Mathematica solves the ODE at this point in time and becauseqis independent ofx. Of course, you may tell Mathematica thatqshould depend onxby usingq[x]instead ofq, but I don't think you will like the result either.
$endgroup$
– Henrik Schumacher
Feb 13 at 13:24
$begingroup$
I don't understand what you mean by "automation". The difference between=and:=lies in the time of execution of the right hand side. If you want that the ODE is solved only after a precise value ofqis supplied, you have to use:=. And yes, there are a lot of questions on the subtle differences of=and:=on this site. And I have to admit that was also something that was quite miraculous to me when I started to use the language. It helps a bit to recall that Mathematica, in its core, is nothing else but a term rewriting system.
$endgroup$
– Henrik Schumacher
Feb 13 at 13:28
|
show 3 more comments
9
9
9
$begingroup$
You attempt to solve the ODE at the time of definition of lsolve. At this point, q does not depend on x. You really want to use SetDelayed here:
lsolve[r_, q_, a_, eta_] :=
y[x] /. First@DSolve[{y'[x] == r - q*y[x], y[a] == eta}, y[x], x];
nwe = lsolve[x, 2*x, 0, 0] // Simplify
1/2 - E^-x^2/2
answered Feb 12 at 15:31
Henrik SchumacherHenrik Schumacher
60.3k583169
$endgroup$
You attempt to solve the ODE at the time of definition of lsolve. At this point, q does not depend on x. You really want to use SetDelayed here:
lsolve[r_, q_, a_, eta_] :=
y[x] /. First@DSolve[{y'[x] == r - q*y[x], y[a] == eta}, y[x], x];
nwe = lsolve[x, 2*x, 0, 0] // Simplify
1/2 - E^-x^2/2
answered Feb 12 at 15:31
Henrik SchumacherHenrik Schumacher
60.3k583169
edited Feb 12 at 20:49
answered Feb 12 at 15:31
Henrik SchumacherHenrik Schumacher
60.3k583169
answered Feb 12 at 15:31
Henrik SchumacherHenrik Schumacher
60.3k583169
answered Feb 12 at 15:31
Henrik SchumacherHenrik Schumacher
60.3k583169
60.3k583169
$begingroup$
Thanks, I'm glad I asked. Even though my version was syntactically correct, it would have been nice if Mathematica produced a warning!
$endgroup$
– Joe Corneli
Feb 12 at 15:41
$begingroup$
How should Mathematica be able to detect that this does not align with your expectations? Both syntaxesf[x_] = ...andf[x_] := ...have the applications, e.g., the former can be considered as a shorthand forf[x_] := Evaluate[...]. This is usefull whenever a symbolic compuation can already performed at the time of definition so that it need not be redone every timefis called.
$endgroup$
– Henrik Schumacher
Feb 12 at 17:17
$begingroup$
As you said in the answer: "At this point,qdoes not depend onx." I think Mathematica should be able to tell that I was building an expression that does depend onxand then generate a warning (but certainly not an error) very much along the lines of what you told me. Otherwise everyone who doesn't know about:=will end up asking a question similar to mine. How to preempt that and automate your answer? :-)
$endgroup$
– Joe Corneli
Feb 13 at 13:18
$begingroup$
When an assignment is made with=,Mathematicacan only set the left hand side to the value of the right hand side at the time of assignment. Evaluatey[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x](just in case: in a fresh kernel) and you will see that the result is(E^(-q x) (E^(a q) eta q - E^(a q) r + E^(q x) r))/q. Because Mathematica solves the ODE at this point in time and becauseqis independent ofx. Of course, you may tell Mathematica thatqshould depend onxby usingq[x]instead ofq, but I don't think you will like the result either.
$endgroup$
– Henrik Schumacher
Feb 13 at 13:24
$begingroup$
I don't understand what you mean by "automation". The difference between=and:=lies in the time of execution of the right hand side. If you want that the ODE is solved only after a precise value ofqis supplied, you have to use:=. And yes, there are a lot of questions on the subtle differences of=and:=on this site. And I have to admit that was also something that was quite miraculous to me when I started to use the language. It helps a bit to recall that Mathematica, in its core, is nothing else but a term rewriting system.
$endgroup$
– Henrik Schumacher
Feb 13 at 13:28
|
show 3 more comments
$begingroup$
Thanks, I'm glad I asked. Even though my version was syntactically correct, it would have been nice if Mathematica produced a warning!
$endgroup$
– Joe Corneli
Feb 12 at 15:41
$begingroup$
How should Mathematica be able to detect that this does not align with your expectations? Both syntaxesf[x_] = ...andf[x_] := ...have the applications, e.g., the former can be considered as a shorthand forf[x_] := Evaluate[...]. This is usefull whenever a symbolic compuation can already performed at the time of definition so that it need not be redone every timefis called.
$endgroup$
– Henrik Schumacher
Feb 12 at 17:17
$begingroup$
As you said in the answer: "At this point,qdoes not depend onx." I think Mathematica should be able to tell that I was building an expression that does depend onxand then generate a warning (but certainly not an error) very much along the lines of what you told me. Otherwise everyone who doesn't know about:=will end up asking a question similar to mine. How to preempt that and automate your answer? :-)
$endgroup$
– Joe Corneli
Feb 13 at 13:18
$begingroup$
When an assignment is made with=,Mathematicacan only set the left hand side to the value of the right hand side at the time of assignment. Evaluatey[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x](just in case: in a fresh kernel) and you will see that the result is(E^(-q x) (E^(a q) eta q - E^(a q) r + E^(q x) r))/q. Because Mathematica solves the ODE at this point in time and becauseqis independent ofx. Of course, you may tell Mathematica thatqshould depend onxby usingq[x]instead ofq, but I don't think you will like the result either.
$endgroup$
– Henrik Schumacher
Feb 13 at 13:24
$begingroup$
I don't understand what you mean by "automation". The difference between=and:=lies in the time of execution of the right hand side. If you want that the ODE is solved only after a precise value ofqis supplied, you have to use:=. And yes, there are a lot of questions on the subtle differences of=and:=on this site. And I have to admit that was also something that was quite miraculous to me when I started to use the language. It helps a bit to recall that Mathematica, in its core, is nothing else but a term rewriting system.
$endgroup$
– Henrik Schumacher
Feb 13 at 13:28
$begingroup$
Thanks, I'm glad I asked. Even though my version was syntactically correct, it would have been nice if Mathematica produced a warning!
$endgroup$
– Joe Corneli
Feb 12 at 15:41
$begingroup$
Thanks, I'm glad I asked. Even though my version was syntactically correct, it would have been nice if Mathematica produced a warning!
$endgroup$
– Joe Corneli
Feb 12 at 15:41
$begingroup$
How should Mathematica be able to detect that this does not align with your expectations? Both syntaxes
f[x_] = ... and f[x_] := ... have the applications, e.g., the former can be considered as a shorthand for f[x_] := Evaluate[...]. This is usefull whenever a symbolic compuation can already performed at the time of definition so that it need not be redone every time f is called.$endgroup$
– Henrik Schumacher
Feb 12 at 17:17
$begingroup$
How should Mathematica be able to detect that this does not align with your expectations? Both syntaxes
f[x_] = ... and f[x_] := ... have the applications, e.g., the former can be considered as a shorthand for f[x_] := Evaluate[...]. This is usefull whenever a symbolic compuation can already performed at the time of definition so that it need not be redone every time f is called.$endgroup$
– Henrik Schumacher
Feb 12 at 17:17
$begingroup$
As you said in the answer: "At this point,
q does not depend on x." I think Mathematica should be able to tell that I was building an expression that does depend on x and then generate a warning (but certainly not an error) very much along the lines of what you told me. Otherwise everyone who doesn't know about := will end up asking a question similar to mine. How to preempt that and automate your answer? :-)$endgroup$
– Joe Corneli
Feb 13 at 13:18
$begingroup$
As you said in the answer: "At this point,
q does not depend on x." I think Mathematica should be able to tell that I was building an expression that does depend on x and then generate a warning (but certainly not an error) very much along the lines of what you told me. Otherwise everyone who doesn't know about := will end up asking a question similar to mine. How to preempt that and automate your answer? :-)$endgroup$
– Joe Corneli
Feb 13 at 13:18
$begingroup$
When an assignment is made with
=, Mathematica can only set the left hand side to the value of the right hand side at the time of assignment. Evaluate y[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x] (just in case: in a fresh kernel) and you will see that the result is (E^(-q x) (E^(a q) eta q - E^(a q) r + E^(q x) r))/q. Because Mathematica solves the ODE at this point in time and because q is independent of x. Of course, you may tell Mathematica that q should depend on x by using q[x] instead of q, but I don't think you will like the result either.$endgroup$
– Henrik Schumacher
Feb 13 at 13:24
$begingroup$
When an assignment is made with
=, Mathematica can only set the left hand side to the value of the right hand side at the time of assignment. Evaluate y[x]/.First@DSolve[{y'[x]==r-q*y[x],y[a]==eta},y[x],x] (just in case: in a fresh kernel) and you will see that the result is (E^(-q x) (E^(a q) eta q - E^(a q) r + E^(q x) r))/q. Because Mathematica solves the ODE at this point in time and because q is independent of x. Of course, you may tell Mathematica that q should depend on x by using q[x] instead of q, but I don't think you will like the result either.$endgroup$
– Henrik Schumacher
Feb 13 at 13:24
$begingroup$
I don't understand what you mean by "automation". The difference between
= and := lies in the time of execution of the right hand side. If you want that the ODE is solved only after a precise value of q is supplied, you have to use :=. And yes, there are a lot of questions on the subtle differences of = and := on this site. And I have to admit that was also something that was quite miraculous to me when I started to use the language. It helps a bit to recall that Mathematica, in its core, is nothing else but a term rewriting system.$endgroup$
– Henrik Schumacher
Feb 13 at 13:28
$begingroup$
I don't understand what you mean by "automation". The difference between
= and := lies in the time of execution of the right hand side. If you want that the ODE is solved only after a precise value of q is supplied, you have to use :=. And yes, there are a lot of questions on the subtle differences of = and := on this site. And I have to admit that was also something that was quite miraculous to me when I started to use the language. It helps a bit to recall that Mathematica, in its core, is nothing else but a term rewriting system.$endgroup$
– Henrik Schumacher
Feb 13 at 13:28
|
show 3 more comments
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$begingroup$
Here are the docs on this issue, in case it helps someone else. reference.wolfram.com/language/tutorial/…
$endgroup$
– Joe Corneli
Feb 12 at 15:42