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# Why I cannot get the results again?

Posted 9 years ago
 Here is my equations, and I solved it once in a few second. But it cannot solve again when I tried to make a change (I did not change yet), (where E^ is for exponentiation) Solve[{2*Dne*n*a1 == a4*Dse, Dse*a4 == Dpe*4*a6*(n + s - l), a1*n^2 + a2 == a4*n + a5, a6*(l - n - s)^2 + a7 == a4*(n + s) + a5, -a1*en*n^3*b*E^(b*t)/3 - a2*n*b*E^(b*t) == a + 2 Dne*n*(1 - E^(b*t)), -a4*es*b*((n + s)^2 - n^2)*E^(b*t)/2 - a5*b*es*s*E^(b* t) == 0, ep*a6*b*E^(b*t)*(n + s - l)/3 - a7*ep*(l - n - s)*E^(b*t) == -a - Dps*4*a6*(1 - E^(b*t))*(n + s - l)}, {a1, a2, a4, a5, a6, a7, b}]  below is the solution it give me at first time: {{a1 -> -(( a Dse (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s))/( Dne ep n^2 (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), a2 -> -((a (Dse n + Dne s) (-Dne Dse ep - 6 Dpe Dse n^2 + 2 Dpe Dse en n^2 - 6 Dne Dpe n s))/( Dne ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), a4 -> -((2 a (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s))/( ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), a5 -> -((a (2 n + s) (-Dne Dse ep - 6 Dpe Dse n^2 + 2 Dpe Dse en n^2 - 6 Dne Dpe n s))/( ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), a6 -> -((a Dse (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s))/( 2 Dpe ep n (-l + n + s)^2 (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), a7 -> (a (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s))/( 2 Dpe ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s)), b -> (3 Dne ep (l - n - s) (Dse l - Dse n + 2 Dpe s - Dse s))/( Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s)}}  I have no idea why the program cannot solve the equation again. Thank you a lot!
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Posted 9 years ago
 The behavior you are seeing was a design decision in the forum software that it was more important to make the default behavior of * be to toggle italic on or off, and hide the * in the process, than it was to display the * as typed. An * separated by white space is exempt from this. Many new users do not realize this.
Posted 9 years ago
 Hi Jie Li,I think something has happened to your original equation set. There is no separation between the variables, so the multiplications have been lost. For example, if you copy out the equations list and then do a Shift/Return, you get {2 Dnena1 == a4Dse, Dsea4 == Dpe4a6 (-l + n + s), a1n^2 + a2 == a4n + a5, a7 + a6 (l - n - s)^2 == a5 + a4 (n + s), -a2nbE^bt - (a1enn^3 bE^bt)/3 == a + 2 Dnen (1 - E^bt), -a5bessE^bt - 1/2 a4esb E^bt (-n^2 + (n + s)^2) == 0, -a7ep E^bt (l - n - s) + 1/3 epa6bE^bt (-l + n + s) == -a - Dps4a6 (1 - E^bt) (-l + n + s)} Eric
Posted 9 years ago
 Thank your reply, I edited the code but it still not work.