Wrong ODE solution in Mathematica 7?
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From: Zsolt on 4 Jan 2010 06:00 Hi! I tried solve the ODE: DSolve[D[y[x], x, x] == -Cos[x]/(1 + Sin[x])^2, y[x], x] The solution what M7 (and Wolfram Alpha) gives is: y[x] -> C[1] + x C[2] + (2 Sin[x/2])/(Cos[x/2] + Sin[x/2]) I think, it's wrong! (Does anybody know how to check?) Another system gives for the same diff.eq: y(x) = -2/(tan((1/2)*x)+1)+_C1*x+_C2 (similar, but not the same->ctan vs tan...) I found the problem in one of my math books, and the solution there concours with the other system. How can I trust Mathematica, if it makes mistakes in such simple things?? :( Thank you for your answer! :)
From: dh on 5 Jan 2010 01:41 Hi, both are correct. You may check this by calculating the second derivative of both expressions and show that they are equal: D[-2/(Tan[(1/2)*x] + 1), {x, 2}] == D[(2 Sin[x/2])/(Cos[x/2] + Sin[x/2]), {x, 2}] // Simplify Daniel Zsolt wrote: > Hi! > I tried solve the ODE: > DSolve[D[y[x], x, x] == -Cos[x]/(1 + Sin[x])^2, y[x], x] > > The solution what M7 (and Wolfram Alpha) gives is: > y[x] -> C[1] + x C[2] + (2 Sin[x/2])/(Cos[x/2] + Sin[x/2]) > > I think, it's wrong! (Does anybody know how to check?) Another system gives > for the same diff.eq: > y(x) = -2/(tan((1/2)*x)+1)+_C1*x+_C2 > (similar, but not the same->ctan vs tan...) > I found the problem in one of my math books, and the solution there > concours with the other system. > How can I trust Mathematica, if it makes mistakes in such simple > things?? :( > Thank you for your answer! :) >
From: Tony Harker on 5 Jan 2010 01:41 Let Mathematica check it: eq = D[y[x], x, x] == -Cos[x]/(1 + Sin[x])^2 sol = DSolve[eq, y, x] eq /. sol[[1]] // Simplify returns True. So the solution is correct, and if you think about it the difference between Mathematica's solution and the other one is only a matter of a difference in the additive constant. Tony ]-> -----Original Message----- ]-> From: Zsolt [mailto:phyhari(a)gmail.com] ]-> Sent: 04 January 2010 10:59 ]-> To: mathgroup(a)smc.vnet.net ]-> Subject: Wrong ODE solution in Mathematica 7? ]-> ]-> Hi! ]-> I tried solve the ODE: ]-> DSolve[D[y[x], x, x] == -Cos[x]/(1 + Sin[x])^2, y[x], x] ]-> ]-> The solution what M7 (and Wolfram Alpha) gives is: ]-> y[x] -> C[1] + x C[2] + (2 Sin[x/2])/(Cos[x/2] + Sin[x/2]) ]-> ]-> I think, it's wrong! (Does anybody know how to check?) ]-> Another system gives for the same diff.eq: ]-> y(x) = -2/(tan((1/2)*x)+1)+_C1*x+_C2 ]-> (similar, but not the same->ctan vs tan...) I found the ]-> problem in one of my math books, and the solution there ]-> concours with the other system. ]-> How can I trust Mathematica, if it makes mistakes in such ]-> simple things?? :( Thank you for your answer! :) ]-> ]->
From: David Park on 5 Jan 2010 01:42 Remember that the constants are arbitrary and so it is possible to obtain different looking forms depending on how you pick the constants. Clear[y] DSolve[y''[x] == -Cos[x]/(1 + Sin[x])^2, y, x][[1, 1]]; y[x_] = y[x] /. % C[1] + x C[2] + (2 Sin[x/2])/(Cos[x/2] + Sin[x/2]) Check the solution: y''[x] == -Cos[x]/(1 + Sin[x])^2 // Simplify True Here is your second solution: y2[x_] = C[1] + C[2] x - 2/(Tan[x/2] + 1); Check it: y2''[x] == -Cos[x]/(1 + Sin[x])^2 // Simplify True Add the constant 2, which could be absorbed into C[1], to the y2 solution and see if it is equal to the y solution: y2[x] + 2 == y[x] // Simplify True How did I figure out how to add 2? By plotting and trying various constants. Start out with zero and you can see that you need to shift the second curve up. Plot[{(2 Sin[x/2])/( Cos[x/2] + Sin[x/2]), -(2/(Tan[x/2] + 1)) + 2}, {x, -2 \[Pi], 2 \[Pi]}] David Park djmpark(a)comcast.net http://home.comcast.net/~djmpark/ From: Zsolt [mailto:phyhari(a)gmail.com] Hi! I tried solve the ODE: DSolve[D[y[x], x, x] == -Cos[x]/(1 + Sin[x])^2, y[x], x] The solution what M7 (and Wolfram Alpha) gives is: y[x] -> C[1] + x C[2] + (2 Sin[x/2])/(Cos[x/2] + Sin[x/2]) I think, it's wrong! (Does anybody know how to check?) Another system gives for the same diff.eq: y(x) = -2/(tan((1/2)*x)+1)+_C1*x+_C2 (similar, but not the same->ctan vs tan...) I found the problem in one of my math books, and the solution there concours with the other system. How can I trust Mathematica, if it makes mistakes in such simple things?? :( Thank you for your answer! :)
From: Kevin J. McCann on 5 Jan 2010 01:43 If you take the second derivative of your answer and compare it with the rhs you will see that it is true, i.e. Y''[x]== -Cos[x]/(1 + Sin[x])^2//FullSimplify Zsolt wrote: > Hi! > I tried solve the ODE: > DSolve[D[y[x], x, x] == -Cos[x]/(1 + Sin[x])^2, y[x], x] > > The solution what M7 (and Wolfram Alpha) gives is: > y[x] -> C[1] + x C[2] + (2 Sin[x/2])/(Cos[x/2] + Sin[x/2]) > > I think, it's wrong! (Does anybody know how to check?) Another system gives > for the same diff.eq: > y(x) = -2/(tan((1/2)*x)+1)+_C1*x+_C2 > (similar, but not the same->ctan vs tan...) > I found the problem in one of my math books, and the solution there > concours with the other system. > How can I trust Mathematica, if it makes mistakes in such simple > things?? :( > Thank you for your answer! :) >
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